The velocity at t = 2 s .

Question

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Answer 1

Question

Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at t=2 s .

(a)

Expert Solution

Answer to Problem 50P

v→(2)=10i^−3j^

Explanation of Solution

Given data

x(0)=4.0 my(0)=3.0 mv→(0)=2i^−9j^a→(t)=4i^+3j^

Formula used

v→(t)=∫a→(t)dt

Calculation

v→(t)=∫ a →( t)dtv→(t)=∫( 4 i ^ +3 j ^ )dt

v→(t)=4i^(t+a)+3j^(t+b) where a and b are constants.

Substitute the values for v→(t) when t=0 s

v→(0)=4i^(a)+3j^(b)=2i^−9j^

Comparing values of i^ ,

4a=2a=0.5

Comparing values of j^ ,

3b=−9b=−3

Substituting values of a and b,

v→(t)=4(t+0.5)i^+3(t−3)j^

when t=2 s

v→(2)=4(2+0.5)i^+3(2−3)j^v→(2)=4(2.5)i^+3(−1)j^v→(2)=10i^−3j^

Conclusion

v→(2)=10i^−3j^

(b)

To determine

To Calculate: The position vector at t=4 s.

(b)

Expert Solution

Answer to Problem 50P

s→(4)=41i^−3j^

Magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s= θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x direction to negative y direction.

Explanation of Solution

Given data

v→(t)=4i^(t+0.5)+3j^(t−3)

Formula used

s→(t)=∫v→(t)dt

Calculation

s→(t)=∫ v →( t)dts→(t)=∫[ 4 i ^ ( t+0.5 )+3 j ^ ( t−3 )]dt

s→(t)=4i^(t22+0.5t+c)+3j^(t22−3t+d) where c and d are constants.

Substituting values for s→(t) when t=0 s

s→(0)=4( 0 2 2+0.5×0+c)i^+3( 0 2 2−3×0+d)j^s→(0)=4ci^+3dj^

Comparing values of i^

t=0 s ,

4c=1c=0.25

Comparing values of j^ when t=0 s ,

3d=3d=1

Substituting values of c and d,

s→(t)=4(t22+0.5t+0.25)i^+3(t22−3t+3)j^

When t=4 s

s→(4)=4( 4 2 2+0.5×4+0.25)i^+3( 4 2 2−3×4+3)j^s→(4)=4( 162+2+0.25)i^+3( 162−12+3)j^s→(4)=4(10.25)i^+3(−1)j^s→(4)=41i^−3j^

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

θ=tan−1(3 41)=tan−1(0.0732)=4.19∘

Conclusion

s→(4)=41i^−3j^

The magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s = θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x-direction to negative y-direction.

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Answer 2

Question

Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at t=2 s .

(a)

Expert Solution

Answer to Problem 50P

v→(2)=10i^−3j^

Explanation of Solution

Given data

x(0)=4.0 my(0)=3.0 mv→(0)=2i^−9j^a→(t)=4i^+3j^

Formula used

v→(t)=∫a→(t)dt

Calculation

v→(t)=∫ a →( t)dtv→(t)=∫( 4 i ^ +3 j ^ )dt

v→(t)=4i^(t+a)+3j^(t+b) where a and b are constants.

Substitute the values for v→(t) when t=0 s

v→(0)=4i^(a)+3j^(b)=2i^−9j^

Comparing values of i^ ,

4a=2a=0.5

Comparing values of j^ ,

3b=−9b=−3

Substituting values of a and b,

v→(t)=4(t+0.5)i^+3(t−3)j^

when t=2 s

v→(2)=4(2+0.5)i^+3(2−3)j^v→(2)=4(2.5)i^+3(−1)j^v→(2)=10i^−3j^

Conclusion

v→(2)=10i^−3j^

(b)

To determine

To Calculate: The position vector at t=4 s.

(b)

Expert Solution

Answer to Problem 50P

s→(4)=41i^−3j^

Magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s= θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x direction to negative y direction.

Explanation of Solution

Given data

v→(t)=4i^(t+0.5)+3j^(t−3)

Formula used

s→(t)=∫v→(t)dt

Calculation

s→(t)=∫ v →( t)dts→(t)=∫[ 4 i ^ ( t+0.5 )+3 j ^ ( t−3 )]dt

s→(t)=4i^(t22+0.5t+c)+3j^(t22−3t+d) where c and d are constants.

Substituting values for s→(t) when t=0 s

s→(0)=4( 0 2 2+0.5×0+c)i^+3( 0 2 2−3×0+d)j^s→(0)=4ci^+3dj^

Comparing values of i^

t=0 s ,

4c=1c=0.25

Comparing values of j^ when t=0 s ,

3d=3d=1

Substituting values of c and d,

s→(t)=4(t22+0.5t+0.25)i^+3(t22−3t+3)j^

When t=4 s

s→(4)=4( 4 2 2+0.5×4+0.25)i^+3( 4 2 2−3×4+3)j^s→(4)=4( 162+2+0.25)i^+3( 162−12+3)j^s→(4)=4(10.25)i^+3(−1)j^s→(4)=41i^−3j^

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

θ=tan−1(3 41)=tan−1(0.0732)=4.19∘

Conclusion

s→(4)=41i^−3j^

The magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s = θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x-direction to negative y-direction.

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Answer 3

Question

Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at t=2 s .

(a)

Expert Solution

Answer to Problem 50P

v→(2)=10i^−3j^

Explanation of Solution

Given data

x(0)=4.0 my(0)=3.0 mv→(0)=2i^−9j^a→(t)=4i^+3j^

Formula used

v→(t)=∫a→(t)dt

Calculation

v→(t)=∫ a →( t)dtv→(t)=∫( 4 i ^ +3 j ^ )dt

v→(t)=4i^(t+a)+3j^(t+b) where a and b are constants.

Substitute the values for v→(t) when t=0 s

v→(0)=4i^(a)+3j^(b)=2i^−9j^

Comparing values of i^ ,

4a=2a=0.5

Comparing values of j^ ,

3b=−9b=−3

Substituting values of a and b,

v→(t)=4(t+0.5)i^+3(t−3)j^

when t=2 s

v→(2)=4(2+0.5)i^+3(2−3)j^v→(2)=4(2.5)i^+3(−1)j^v→(2)=10i^−3j^

Conclusion

v→(2)=10i^−3j^

(b)

To determine

To Calculate: The position vector at t=4 s.

(b)

Expert Solution

Answer to Problem 50P

s→(4)=41i^−3j^

Magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s= θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x direction to negative y direction.

Explanation of Solution

Given data

v→(t)=4i^(t+0.5)+3j^(t−3)

Formula used

s→(t)=∫v→(t)dt

Calculation

s→(t)=∫ v →( t)dts→(t)=∫[ 4 i ^ ( t+0.5 )+3 j ^ ( t−3 )]dt

s→(t)=4i^(t22+0.5t+c)+3j^(t22−3t+d) where c and d are constants.

Substituting values for s→(t) when t=0 s

s→(0)=4( 0 2 2+0.5×0+c)i^+3( 0 2 2−3×0+d)j^s→(0)=4ci^+3dj^

Comparing values of i^

t=0 s ,

4c=1c=0.25

Comparing values of j^ when t=0 s ,

3d=3d=1

Substituting values of c and d,

s→(t)=4(t22+0.5t+0.25)i^+3(t22−3t+3)j^

When t=4 s

s→(4)=4( 4 2 2+0.5×4+0.25)i^+3( 4 2 2−3×4+3)j^s→(4)=4( 162+2+0.25)i^+3( 162−12+3)j^s→(4)=4(10.25)i^+3(−1)j^s→(4)=41i^−3j^

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

θ=tan−1(3 41)=tan−1(0.0732)=4.19∘

Conclusion

s→(4)=41i^−3j^

The magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s = θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x-direction to negative y-direction.

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Answer 4

Question

Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at t=2 s .

(a)

Expert Solution

Answer to Problem 50P

v→(2)=10i^−3j^

Explanation of Solution

Given data

x(0)=4.0 my(0)=3.0 mv→(0)=2i^−9j^a→(t)=4i^+3j^

Formula used

v→(t)=∫a→(t)dt

Calculation

v→(t)=∫ a →( t)dtv→(t)=∫( 4 i ^ +3 j ^ )dt

v→(t)=4i^(t+a)+3j^(t+b) where a and b are constants.

Substitute the values for v→(t) when t=0 s

v→(0)=4i^(a)+3j^(b)=2i^−9j^

Comparing values of i^ ,

4a=2a=0.5

Comparing values of j^ ,

3b=−9b=−3

Substituting values of a and b,

v→(t)=4(t+0.5)i^+3(t−3)j^

when t=2 s

v→(2)=4(2+0.5)i^+3(2−3)j^v→(2)=4(2.5)i^+3(−1)j^v→(2)=10i^−3j^

Conclusion

v→(2)=10i^−3j^

(b)

To determine

To Calculate: The position vector at t=4 s.

(b)

Expert Solution

Answer to Problem 50P

s→(4)=41i^−3j^

Magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s= θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x direction to negative y direction.

Explanation of Solution

Given data

v→(t)=4i^(t+0.5)+3j^(t−3)

Formula used

s→(t)=∫v→(t)dt

Calculation

s→(t)=∫ v →( t)dts→(t)=∫[ 4 i ^ ( t+0.5 )+3 j ^ ( t−3 )]dt

s→(t)=4i^(t22+0.5t+c)+3j^(t22−3t+d) where c and d are constants.

Substituting values for s→(t) when t=0 s

s→(0)=4( 0 2 2+0.5×0+c)i^+3( 0 2 2−3×0+d)j^s→(0)=4ci^+3dj^

Comparing values of i^

t=0 s ,

4c=1c=0.25

Comparing values of j^ when t=0 s ,

3d=3d=1

Substituting values of c and d,

s→(t)=4(t22+0.5t+0.25)i^+3(t22−3t+3)j^

When t=4 s

s→(4)=4( 4 2 2+0.5×4+0.25)i^+3( 4 2 2−3×4+3)j^s→(4)=4( 162+2+0.25)i^+3( 162−12+3)j^s→(4)=4(10.25)i^+3(−1)j^s→(4)=41i^−3j^

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

θ=tan−1(3 41)=tan−1(0.0732)=4.19∘

Conclusion

s→(4)=41i^−3j^

The magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s = θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x-direction to negative y-direction.

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Answer 5

Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at t=2 s .

(a)

Expert Solution

Answer to Problem 50P

v→(2)=10i^−3j^

Explanation of Solution

Given data

x(0)=4.0 my(0)=3.0 mv→(0)=2i^−9j^a→(t)=4i^+3j^

Formula used

v→(t)=∫a→(t)dt

Calculation

v→(t)=∫ a →( t)dtv→(t)=∫( 4 i ^ +3 j ^ )dt

v→(t)=4i^(t+a)+3j^(t+b) where a and b are constants.

Substitute the values for v→(t) when t=0 s

v→(0)=4i^(a)+3j^(b)=2i^−9j^

Comparing values of i^ ,

4a=2a=0.5

Comparing values of j^ ,

3b=−9b=−3

Substituting values of a and b,

v→(t)=4(t+0.5)i^+3(t−3)j^

when t=2 s

v→(2)=4(2+0.5)i^+3(2−3)j^v→(2)=4(2.5)i^+3(−1)j^v→(2)=10i^−3j^

Conclusion

v→(2)=10i^−3j^

(b)

To determine

To Calculate: The position vector at t=4 s.

(b)

Expert Solution

Answer to Problem 50P

s→(4)=41i^−3j^

Magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s= θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x direction to negative y direction.

Explanation of Solution

Given data

v→(t)=4i^(t+0.5)+3j^(t−3)

Formula used

s→(t)=∫v→(t)dt

Calculation

s→(t)=∫ v →( t)dts→(t)=∫[ 4 i ^ ( t+0.5 )+3 j ^ ( t−3 )]dt

s→(t)=4i^(t22+0.5t+c)+3j^(t22−3t+d) where c and d are constants.

Substituting values for s→(t) when t=0 s

s→(0)=4( 0 2 2+0.5×0+c)i^+3( 0 2 2−3×0+d)j^s→(0)=4ci^+3dj^

Comparing values of i^

t=0 s ,

4c=1c=0.25

Comparing values of j^ when t=0 s ,

3d=3d=1

Substituting values of c and d,

s→(t)=4(t22+0.5t+0.25)i^+3(t22−3t+3)j^

When t=4 s

s→(4)=4( 4 2 2+0.5×4+0.25)i^+3( 4 2 2−3×4+3)j^s→(4)=4( 162+2+0.25)i^+3( 162−12+3)j^s→(4)=4(10.25)i^+3(−1)j^s→(4)=41i^−3j^

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

θ=tan−1(3 41)=tan−1(0.0732)=4.19∘

Conclusion

s→(4)=41i^−3j^

The magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s = θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x-direction to negative y-direction.

Answer 6

Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at t=2 s .

(a)

Expert Solution

Answer to Problem 50P

v→(2)=10i^−3j^

Explanation of Solution

Given data

x(0)=4.0 my(0)=3.0 mv→(0)=2i^−9j^a→(t)=4i^+3j^

Formula used

v→(t)=∫a→(t)dt

Calculation

v→(t)=∫ a →( t)dtv→(t)=∫( 4 i ^ +3 j ^ )dt

v→(t)=4i^(t+a)+3j^(t+b) where a and b are constants.

Substitute the values for v→(t) when t=0 s

v→(0)=4i^(a)+3j^(b)=2i^−9j^

Comparing values of i^ ,

4a=2a=0.5

Comparing values of j^ ,

3b=−9b=−3

Substituting values of a and b,

v→(t)=4(t+0.5)i^+3(t−3)j^

when t=2 s

v→(2)=4(2+0.5)i^+3(2−3)j^v→(2)=4(2.5)i^+3(−1)j^v→(2)=10i^−3j^

Conclusion

v→(2)=10i^−3j^

Answer 7

Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at t=2 s .

Answer 8

(a)

Expert Solution

Answer to Problem 50P

v→(2)=10i^−3j^

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given data

x(0)=4.0 my(0)=3.0 mv→(0)=2i^−9j^a→(t)=4i^+3j^

Formula used

v→(t)=∫a→(t)dt

Calculation

v→(t)=∫ a →( t)dtv→(t)=∫( 4 i ^ +3 j ^ )dt

v→(t)=4i^(t+a)+3j^(t+b) where a and b are constants.

Substitute the values for v→(t) when t=0 s

v→(0)=4i^(a)+3j^(b)=2i^−9j^

Comparing values of i^ ,

4a=2a=0.5

Comparing values of j^ ,

3b=−9b=−3

Substituting values of a and b,

v→(t)=4(t+0.5)i^+3(t−3)j^

when t=2 s

v→(2)=4(2+0.5)i^+3(2−3)j^v→(2)=4(2.5)i^+3(−1)j^v→(2)=10i^−3j^

Conclusion

v→(2)=10i^−3j^

Answer 13

(b)

To determine

To Calculate: The position vector at t=4 s.

(b)

Expert Solution

Answer to Problem 50P

s→(4)=41i^−3j^

Magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s= θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x direction to negative y direction.

Explanation of Solution

Given data

v→(t)=4i^(t+0.5)+3j^(t−3)

Formula used

s→(t)=∫v→(t)dt

Calculation

s→(t)=∫ v →( t)dts→(t)=∫[ 4 i ^ ( t+0.5 )+3 j ^ ( t−3 )]dt

s→(t)=4i^(t22+0.5t+c)+3j^(t22−3t+d) where c and d are constants.

Substituting values for s→(t) when t=0 s

s→(0)=4( 0 2 2+0.5×0+c)i^+3( 0 2 2−3×0+d)j^s→(0)=4ci^+3dj^

Comparing values of i^

t=0 s ,

4c=1c=0.25

Comparing values of j^ when t=0 s ,

3d=3d=1

Substituting values of c and d,

s→(t)=4(t22+0.5t+0.25)i^+3(t22−3t+3)j^

When t=4 s

s→(4)=4( 4 2 2+0.5×4+0.25)i^+3( 4 2 2−3×4+3)j^s→(4)=4( 162+2+0.25)i^+3( 162−12+3)j^s→(4)=4(10.25)i^+3(−1)j^s→(4)=41i^−3j^

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

θ=tan−1(3 41)=tan−1(0.0732)=4.19∘

Conclusion

s→(4)=41i^−3j^

The magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s = θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x-direction to negative y-direction.

Answer 14

(b)

To determine

To Calculate: The position vector at t=4 s.

Answer 15

(b)

Expert Solution

Answer to Problem 50P

s→(4)=41i^−3j^

Magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s= θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x direction to negative y direction.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given data

v→(t)=4i^(t+0.5)+3j^(t−3)

Formula used

s→(t)=∫v→(t)dt

Calculation

s→(t)=∫ v →( t)dts→(t)=∫[ 4 i ^ ( t+0.5 )+3 j ^ ( t−3 )]dt

s→(t)=4i^(t22+0.5t+c)+3j^(t22−3t+d) where c and d are constants.

Substituting values for s→(t) when t=0 s

s→(0)=4( 0 2 2+0.5×0+c)i^+3( 0 2 2−3×0+d)j^s→(0)=4ci^+3dj^

Comparing values of i^

t=0 s ,

4c=1c=0.25

Comparing values of j^ when t=0 s ,

3d=3d=1

Substituting values of c and d,

s→(t)=4(t22+0.5t+0.25)i^+3(t22−3t+3)j^

When t=4 s

s→(4)=4( 4 2 2+0.5×4+0.25)i^+3( 4 2 2−3×4+3)j^s→(4)=4( 162+2+0.25)i^+3( 162−12+3)j^s→(4)=4(10.25)i^+3(−1)j^s→(4)=41i^−3j^

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

θ=tan−1(3 41)=tan−1(0.0732)=4.19∘

Conclusion

s→(4)=41i^−3j^

The magnitude of the displacement at t=4 s= |s→(4)|

|s→(4)|=412+( −3)2=1681+9=1690=6.411 m

Direction of the displacement at t=4 s = θ

θ=tan−1(341)=tan−1(0.0732)=4.19∘ from positive x-direction to negative y-direction.

Answer 20

Ch. 3 - Prob. 1P Ch. 3 - Prob. 2P Ch. 3 - Prob. 3P Ch. 3 - Prob. 4P Ch. 3 - Prob. 5P Ch. 3 - Prob. 6P Ch. 3 - Prob. 7P Ch. 3 - Prob. 8P Ch. 3 - Prob. 9P Ch. 3 - Prob. 10P

The velocity at t = 2 s .

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To Calculate: The velocity at t=2 s .

Answer to Problem 50P

Explanation of Solution

To Calculate: The position vector at t=4 s.

Answer to Problem 50P

Explanation of Solution

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Chapter 3 Solutions

The velocity at t = 2 ​ s .

Concept explainers

Videos

To Calculate: The velocity at t=2​ s .

Answer to Problem 50P

Explanation of Solution

To Calculate: The position vector at t=4​ s.

Answer to Problem 50P

Explanation of Solution

Want to see more full solutions like this?

Chapter 3 Solutions

The velocity at t = 2 s .

To Calculate: The velocity at t=2 s .

To Calculate: The position vector at t=4 s.