Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 3, Problem 49P

(a)

To determine

To Calculate: The magnitude and the direction of Δvav during the 3.0 s interval.

(a)

Expert Solution
Check Mark

Answer to Problem 49P

Magnitude of the resultant velocity vector Δvav is 36.06 ms-1.

Direction of the resultant velocity vector Δvav is 47.15 .

Explanation of Solution

Given data

Velocity at t = 0 s = 40 m.s-1

The angle of velocity at t = 0 s = θ=45

Velocity at t = 3 s = 30 m.s-1

The angle of velocity at t =3 s =θ=50

Displacement of x at t =3 s=100m

Displacement of y at t =3 s = 80m

Formula used

  Physics for Scientists and Engineers, Chapter 3, Problem 49P

  |v|= | v x |2+ | v y |2θ=tan1( | v y | | v x |)

  |v| denotes the magnitude of the vector and θ denotes the direction of the vector v

  a=ΔvΔt

where a is the acceleration and v and t represent velocity and time respectively.

Calculation

At t=0 s .

Velocity component in horizontal direction = 40 m.s-1×cos45=40 m.s-1×12=28.284 m.s-1

Velocity component in vertical direction = 40 m.s-1×sin45=40 m.s-1×12=28.284 m.s-1

At t=3.0 s ,

Velocity component in horizontal direction = 30 m.s-1×cos50=30 m.s-1×0.6428=19.284 m.s-1

Velocity component in vertical direction = 30 m.s-1×sin50=30 m.s-1×0.7660=22.98 m.s-1

Total velocity components in horizontal direction = 28.284 m.s-1+19.284 m.s-1=47.568 m.s-1

Total velocity components in vertical direction = 28.284 m.s-1+22.98 m.s-1=51.264 m.s-1

Resultant velocity after t=3 s = 47.5682+51.2642 m.s-1=69.93 m.s-1

  θ=tan1(51.26447.568)=tan1(1.078)=47.15

Conclusion

The magnitude of the resultant velocity vector Δvav = 36.06 ms-1

The direction of the resultant velocity vector Δvav = 47.15 from the horizontal direction to vertical direction.

(b)

To determine

To Calculate: The magnitude and the direction of aav .

(b)

Expert Solution
Check Mark

Answer to Problem 49P

Magnitude of the resultant acceleration vector aav is 3.482ms-2

Direction of the resultant acceleration vector aav is. 30.50 from negative x direction to negative y direction

Explanation of Solution

Given data

Velocity at t = 0 s = 40 m.s-1

The angle of velocity at t = 0 s = θ=45

Velocity at t = 3 s =30 m.s-1

The angle of velocity at t =3 s = θ=50

Displacement of x at t =3 s =100m

Displacement of y at t =3 s = 80m

Formula used

  a=ΔvΔt

  |aav|= | a x |2+ | a y |2θ=tan1( | a y | | a x |)

Calculation

  a=ΔvΔtax=19.28428.2843 ms-2ax=93 ms-2ax=3 ms-2ay=22.9828.2843 ms-2ay=5.3043 ms-2ay=1.768 ms-2

  |aav|= | a x |2+ | a y |2|aav|= ( 3 )2+ ( 1.768 )2ms-2|aav|=9+3.126ms-2|aav|=12.126ms-2|aav|=3.482ms-2θ=tan1( | a y | | a x |)θ=tan1( 1.768 3)θ=tan1(0.589)θ=30.50

The magnitude of the resultant acceleration vector aav = 3.482ms-2

The direction of the resultant acceleration vector aav = 30.50 from negative x-direction to negative y-direction

Conclusion

Magnitude of the resultant acceleration vector aav is 3.482ms-2

Direction of the resultant acceleration vector aav is 30.50 from negative x direction to negative y direction.

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Chapter 3 Solutions

Physics for Scientists and Engineers

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