Coordinates ( x, t ) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Question

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Answer 1

Question

Chapter 2, Problem 30P

To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Expert Solution & Answer

Answer to Problem 30P

The coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K (x1, t1) and (x′1, t′1) in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K (x2, t2) and (x′2, t′2) in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus (y= y′) and (z= z′)

In system K coordinates at time of bullet firing are as follows:

x1=1 m, t1=3 ns (I)

In system K coordinates at time of target attacking are as follows:

x2=1 m+120 m =121 m (II)

And

t2=t1+distnace tarvel by bullet speed of the bullet

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

t2=3×10−9 s+120 m0.98×3×108 m / s =3×10−9 s+408×10−9 s=411 ns (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

t′1=γ(t1−vx1c2) t′2=γ(t2−vx2c2)x′1=γ(x1−vt1)x′2=γ(x2−vt2) (IV)

Where, γ is the relativistic factor,

y=11−v2/c2

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute v=0.8c in y=11−v2/c2 to get relativistic factor. Thus

y=11−(0.8c)2/c2 =11−0.64=106=53 (V)

Substitute t1=3 ns, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′1=γ(t1−vx1c2) to get t′1. Thus,

t′1=53(3 ns −0.8c×1 mc2) =53(3 ns −0.8×1 m3×108 m / s) =53(3 ns −2.67×10−9 s)=53(3 ns −2.67 ns)

Simplify the above equation

t′1=53(3 ns −2.67 ns) =53×0.33 ns=0.55 ns

Substitute t2=411 ns, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′2=γ(t2−vx2c2) to get t′2. Thus,

t′2=53(411 ns −0.8c×121 mc2) =53(411 ns −0.8×121 m3×108 m / s) =53(411 ns −322.67×10−9 s)=53(411 ns −322.67 ns)

Simplify the above equation

t′1=53(411 ns −322.67 ns)=53×88.33 ns=147.2 ns

Substitute t1=3 ns=3 ×10−9 s, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′1=γ(x1−vt1) to get x′1. Thus,

x′1=53(1 m−0.8×3×108 m / s×3×10−9 s)=53(1 m−0.72 m)=0.47 m

Substitute t2=411 ns=411×10−9 s, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′2=γ(x2−vt2) to get x′2. Thus,

x′2=53(121 m−0.8×3×108 m / s×411×10−9 s)=53(121 m−98.64 m)=37.3 m

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

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Answer 2

Question

Chapter 2, Problem 30P

To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Expert Solution & Answer

Answer to Problem 30P

The coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K (x1, t1) and (x′1, t′1) in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K (x2, t2) and (x′2, t′2) in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus (y= y′) and (z= z′)

In system K coordinates at time of bullet firing are as follows:

x1=1 m, t1=3 ns (I)

In system K coordinates at time of target attacking are as follows:

x2=1 m+120 m =121 m (II)

And

t2=t1+distnace tarvel by bullet speed of the bullet

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

t2=3×10−9 s+120 m0.98×3×108 m / s =3×10−9 s+408×10−9 s=411 ns (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

t′1=γ(t1−vx1c2) t′2=γ(t2−vx2c2)x′1=γ(x1−vt1)x′2=γ(x2−vt2) (IV)

Where, γ is the relativistic factor,

y=11−v2/c2

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute v=0.8c in y=11−v2/c2 to get relativistic factor. Thus

y=11−(0.8c)2/c2 =11−0.64=106=53 (V)

Substitute t1=3 ns, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′1=γ(t1−vx1c2) to get t′1. Thus,

t′1=53(3 ns −0.8c×1 mc2) =53(3 ns −0.8×1 m3×108 m / s) =53(3 ns −2.67×10−9 s)=53(3 ns −2.67 ns)

Simplify the above equation

t′1=53(3 ns −2.67 ns) =53×0.33 ns=0.55 ns

Substitute t2=411 ns, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′2=γ(t2−vx2c2) to get t′2. Thus,

t′2=53(411 ns −0.8c×121 mc2) =53(411 ns −0.8×121 m3×108 m / s) =53(411 ns −322.67×10−9 s)=53(411 ns −322.67 ns)

Simplify the above equation

t′1=53(411 ns −322.67 ns)=53×88.33 ns=147.2 ns

Substitute t1=3 ns=3 ×10−9 s, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′1=γ(x1−vt1) to get x′1. Thus,

x′1=53(1 m−0.8×3×108 m / s×3×10−9 s)=53(1 m−0.72 m)=0.47 m

Substitute t2=411 ns=411×10−9 s, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′2=γ(x2−vt2) to get x′2. Thus,

x′2=53(121 m−0.8×3×108 m / s×411×10−9 s)=53(121 m−98.64 m)=37.3 m

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

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Answer 3

Question

Chapter 2, Problem 30P

To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Expert Solution & Answer

Answer to Problem 30P

The coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K (x1, t1) and (x′1, t′1) in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K (x2, t2) and (x′2, t′2) in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus (y= y′) and (z= z′)

In system K coordinates at time of bullet firing are as follows:

x1=1 m, t1=3 ns (I)

In system K coordinates at time of target attacking are as follows:

x2=1 m+120 m =121 m (II)

And

t2=t1+distnace tarvel by bullet speed of the bullet

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

t2=3×10−9 s+120 m0.98×3×108 m / s =3×10−9 s+408×10−9 s=411 ns (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

t′1=γ(t1−vx1c2) t′2=γ(t2−vx2c2)x′1=γ(x1−vt1)x′2=γ(x2−vt2) (IV)

Where, γ is the relativistic factor,

y=11−v2/c2

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute v=0.8c in y=11−v2/c2 to get relativistic factor. Thus

y=11−(0.8c)2/c2 =11−0.64=106=53 (V)

Substitute t1=3 ns, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′1=γ(t1−vx1c2) to get t′1. Thus,

t′1=53(3 ns −0.8c×1 mc2) =53(3 ns −0.8×1 m3×108 m / s) =53(3 ns −2.67×10−9 s)=53(3 ns −2.67 ns)

Simplify the above equation

t′1=53(3 ns −2.67 ns) =53×0.33 ns=0.55 ns

Substitute t2=411 ns, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′2=γ(t2−vx2c2) to get t′2. Thus,

t′2=53(411 ns −0.8c×121 mc2) =53(411 ns −0.8×121 m3×108 m / s) =53(411 ns −322.67×10−9 s)=53(411 ns −322.67 ns)

Simplify the above equation

t′1=53(411 ns −322.67 ns)=53×88.33 ns=147.2 ns

Substitute t1=3 ns=3 ×10−9 s, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′1=γ(x1−vt1) to get x′1. Thus,

x′1=53(1 m−0.8×3×108 m / s×3×10−9 s)=53(1 m−0.72 m)=0.47 m

Substitute t2=411 ns=411×10−9 s, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′2=γ(x2−vt2) to get x′2. Thus,

x′2=53(121 m−0.8×3×108 m / s×411×10−9 s)=53(121 m−98.64 m)=37.3 m

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

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Answer 4

Question

Chapter 2, Problem 30P

To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Expert Solution & Answer

Answer to Problem 30P

The coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K (x1, t1) and (x′1, t′1) in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K (x2, t2) and (x′2, t′2) in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus (y= y′) and (z= z′)

In system K coordinates at time of bullet firing are as follows:

x1=1 m, t1=3 ns (I)

In system K coordinates at time of target attacking are as follows:

x2=1 m+120 m =121 m (II)

And

t2=t1+distnace tarvel by bullet speed of the bullet

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

t2=3×10−9 s+120 m0.98×3×108 m / s =3×10−9 s+408×10−9 s=411 ns (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

t′1=γ(t1−vx1c2) t′2=γ(t2−vx2c2)x′1=γ(x1−vt1)x′2=γ(x2−vt2) (IV)

Where, γ is the relativistic factor,

y=11−v2/c2

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute v=0.8c in y=11−v2/c2 to get relativistic factor. Thus

y=11−(0.8c)2/c2 =11−0.64=106=53 (V)

Substitute t1=3 ns, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′1=γ(t1−vx1c2) to get t′1. Thus,

t′1=53(3 ns −0.8c×1 mc2) =53(3 ns −0.8×1 m3×108 m / s) =53(3 ns −2.67×10−9 s)=53(3 ns −2.67 ns)

Simplify the above equation

t′1=53(3 ns −2.67 ns) =53×0.33 ns=0.55 ns

Substitute t2=411 ns, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′2=γ(t2−vx2c2) to get t′2. Thus,

t′2=53(411 ns −0.8c×121 mc2) =53(411 ns −0.8×121 m3×108 m / s) =53(411 ns −322.67×10−9 s)=53(411 ns −322.67 ns)

Simplify the above equation

t′1=53(411 ns −322.67 ns)=53×88.33 ns=147.2 ns

Substitute t1=3 ns=3 ×10−9 s, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′1=γ(x1−vt1) to get x′1. Thus,

x′1=53(1 m−0.8×3×108 m / s×3×10−9 s)=53(1 m−0.72 m)=0.47 m

Substitute t2=411 ns=411×10−9 s, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′2=γ(x2−vt2) to get x′2. Thus,

x′2=53(121 m−0.8×3×108 m / s×411×10−9 s)=53(121 m−98.64 m)=37.3 m

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

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Answer 5

Chapter 2, Problem 30P

To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Expert Solution & Answer

Answer to Problem 30P

The coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K (x1, t1) and (x′1, t′1) in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K (x2, t2) and (x′2, t′2) in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus (y= y′) and (z= z′)

In system K coordinates at time of bullet firing are as follows:

x1=1 m, t1=3 ns (I)

In system K coordinates at time of target attacking are as follows:

x2=1 m+120 m =121 m (II)

And

t2=t1+distnace tarvel by bullet speed of the bullet

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

t2=3×10−9 s+120 m0.98×3×108 m / s =3×10−9 s+408×10−9 s=411 ns (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

t′1=γ(t1−vx1c2) t′2=γ(t2−vx2c2)x′1=γ(x1−vt1)x′2=γ(x2−vt2) (IV)

Where, γ is the relativistic factor,

y=11−v2/c2

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute v=0.8c in y=11−v2/c2 to get relativistic factor. Thus

y=11−(0.8c)2/c2 =11−0.64=106=53 (V)

Substitute t1=3 ns, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′1=γ(t1−vx1c2) to get t′1. Thus,

t′1=53(3 ns −0.8c×1 mc2) =53(3 ns −0.8×1 m3×108 m / s) =53(3 ns −2.67×10−9 s)=53(3 ns −2.67 ns)

Simplify the above equation

t′1=53(3 ns −2.67 ns) =53×0.33 ns=0.55 ns

Substitute t2=411 ns, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′2=γ(t2−vx2c2) to get t′2. Thus,

t′2=53(411 ns −0.8c×121 mc2) =53(411 ns −0.8×121 m3×108 m / s) =53(411 ns −322.67×10−9 s)=53(411 ns −322.67 ns)

Simplify the above equation

t′1=53(411 ns −322.67 ns)=53×88.33 ns=147.2 ns

Substitute t1=3 ns=3 ×10−9 s, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′1=γ(x1−vt1) to get x′1. Thus,

x′1=53(1 m−0.8×3×108 m / s×3×10−9 s)=53(1 m−0.72 m)=0.47 m

Substitute t2=411 ns=411×10−9 s, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′2=γ(x2−vt2) to get x′2. Thus,

x′2=53(121 m−0.8×3×108 m / s×411×10−9 s)=53(121 m−98.64 m)=37.3 m

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Answer 6

Chapter 2, Problem 30P

To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Expert Solution & Answer

Answer to Problem 30P

The coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K (x1, t1) and (x′1, t′1) in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K (x2, t2) and (x′2, t′2) in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus (y= y′) and (z= z′)

In system K coordinates at time of bullet firing are as follows:

x1=1 m, t1=3 ns (I)

In system K coordinates at time of target attacking are as follows:

x2=1 m+120 m =121 m (II)

And

t2=t1+distnace tarvel by bullet speed of the bullet

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

t2=3×10−9 s+120 m0.98×3×108 m / s =3×10−9 s+408×10−9 s=411 ns (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

t′1=γ(t1−vx1c2) t′2=γ(t2−vx2c2)x′1=γ(x1−vt1)x′2=γ(x2−vt2) (IV)

Where, γ is the relativistic factor,

y=11−v2/c2

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute v=0.8c in y=11−v2/c2 to get relativistic factor. Thus

y=11−(0.8c)2/c2 =11−0.64=106=53 (V)

Substitute t1=3 ns, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′1=γ(t1−vx1c2) to get t′1. Thus,

t′1=53(3 ns −0.8c×1 mc2) =53(3 ns −0.8×1 m3×108 m / s) =53(3 ns −2.67×10−9 s)=53(3 ns −2.67 ns)

Simplify the above equation

t′1=53(3 ns −2.67 ns) =53×0.33 ns=0.55 ns

Substitute t2=411 ns, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′2=γ(t2−vx2c2) to get t′2. Thus,

t′2=53(411 ns −0.8c×121 mc2) =53(411 ns −0.8×121 m3×108 m / s) =53(411 ns −322.67×10−9 s)=53(411 ns −322.67 ns)

Simplify the above equation

t′1=53(411 ns −322.67 ns)=53×88.33 ns=147.2 ns

Substitute t1=3 ns=3 ×10−9 s, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′1=γ(x1−vt1) to get x′1. Thus,

x′1=53(1 m−0.8×3×108 m / s×3×10−9 s)=53(1 m−0.72 m)=0.47 m

Substitute t2=411 ns=411×10−9 s, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′2=γ(x2−vt2) to get x′2. Thus,

x′2=53(121 m−0.8×3×108 m / s×411×10−9 s)=53(121 m−98.64 m)=37.3 m

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Answer 7

Chapter 2, Problem 30P

To determine

Coordinates (x, t) in System K and Kʹ at time of bullet firing and time when bullet reached at target.

Answer 8

Expert Solution & Answer

Answer to Problem 30P

The coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Let’s coordinates at the timing of bullet firing are as follows:

In system K (x1, t1) and (x′1, t′1) in system Kʹ.

Let’s coordinates at the timing of target attack are as follows:

In system K (x2, t2) and (x′2, t′2) in system Kʹ.

System Kʹ is moving along the x-axis of the system K, therefore coordinates on x-axis in both system will change. Thus (y= y′) and (z= z′)

In system K coordinates at time of bullet firing are as follows:

x1=1 m, t1=3 ns (I)

In system K coordinates at time of target attacking are as follows:

x2=1 m+120 m =121 m (II)

And

t2=t1+distnace tarvel by bullet speed of the bullet

Substitute the distance travelled equal to 120 m and speed of the bullet 0.98c in above equation to get the time coordinate at time of target attack in system K. Thus,

t2=3×10−9 s+120 m0.98×3×108 m / s =3×10−9 s+408×10−9 s=411 ns (III)

Use Lorentz transformations to get the coordinates in the system Kʹ. Thus,

t′1=γ(t1−vx1c2) t′2=γ(t2−vx2c2)x′1=γ(x1−vt1)x′2=γ(x2−vt2) (IV)

Where, γ is the relativistic factor,

y=11−v2/c2

Where, v is the velocity of the system Kʹ and c is the velocity of the light.

Substitute v=0.8c in y=11−v2/c2 to get relativistic factor. Thus

y=11−(0.8c)2/c2 =11−0.64=106=53 (V)

Substitute t1=3 ns, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′1=γ(t1−vx1c2) to get t′1. Thus,

t′1=53(3 ns −0.8c×1 mc2) =53(3 ns −0.8×1 m3×108 m / s) =53(3 ns −2.67×10−9 s)=53(3 ns −2.67 ns)

Simplify the above equation

t′1=53(3 ns −2.67 ns) =53×0.33 ns=0.55 ns

Substitute t2=411 ns, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation t′2=γ(t2−vx2c2) to get t′2. Thus,

t′2=53(411 ns −0.8c×121 mc2) =53(411 ns −0.8×121 m3×108 m / s) =53(411 ns −322.67×10−9 s)=53(411 ns −322.67 ns)

Simplify the above equation

t′1=53(411 ns −322.67 ns)=53×88.33 ns=147.2 ns

Substitute t1=3 ns=3 ×10−9 s, x1=1 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′1=γ(x1−vt1) to get x′1. Thus,

x′1=53(1 m−0.8×3×108 m / s×3×10−9 s)=53(1 m−0.72 m)=0.47 m

Substitute t2=411 ns=411×10−9 s, x2=121 m, γ=53, c=3×108 m / s, and v=0.8c in equation x′2=γ(x2−vt2) to get x′2. Thus,

x′2=53(121 m−0.8×3×108 m / s×411×10−9 s)=53(121 m−98.64 m)=37.3 m

Conclusion:

Therefore, the coordinates in system K at time of bullet firing is (1 m, 3 ns) and at time of target attack is (121 m, 411 ns) and in system Kʹ are (0.47 m, 0.55 ns) and (37.3 m, 147.2 ns) respectively.