PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 3, Problem 50P

(a)

To determine

To Calculate: The velocity at t=2 s .

(a)

Expert Solution
Check Mark

Answer to Problem 50P

  v(2)=10i^3j^

Explanation of Solution

Given data

  x(0)=4.0my(0)=3.0mv(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(t3)j^

when t=2 s

  v(2)=4(2+0.5)i^+3(23)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
Check Mark

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=tan1(341)=tan1(0.0732)=4.19 from positive x direction to negative y direction.

Explanation of Solution

Given data

  v(t)=4i^(t+0.5)+3j^(t3)

Formula used

  s(t)=v(t)dt

Calculation

  s(t)= v ( t)dts(t)=[ 4 i ^ ( t+0.5 )+3 j ^ ( t3 )]dt

  s(t)=4i^(t22+0.5t+c)+3j^(t223t+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 23×0+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(t223t+3)j^

When t=4 s

  s(4)=4( 4 2 2+0.5×4+0.25)i^+3( 4 2 23×4+3)j^s(4)=4( 162+2+0.25)i^+3( 16212+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=tan1(3 41)=tan1(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=tan1(341)=tan1(0.0732)=4.19 from positive x-direction to negative y-direction.

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Chapter 3 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

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