A ball starts from rest and accelerates at 0.525 m/s^2 while moving down an incl

Question

A ball starts from rest and accelerates at 0.525 m/s^2 while moving down an inclined plane 9.95 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 15.90 m, it comes to rest. (Assume positive direction points down the first plane and up the second plane.) (a) What is the speed of the ball at the bottom of the first plane? (e) Draw position, velocity, and acceleration graphs. These must be turned in on paper. A motorist drives along a straight road at a constant speed of 20.0 m/s. Just as she overtake her. Assume that the officer maintains this acceleration. 0.50M/s^2 to overtake her assume that the officer maintains this acceleration. (a) Determine the time it takes the police officer to reach the motorist. (d) Draw position, velocity, and acceleration graphs for the motorist and officer. You should have 3 graphs, one for position (with both motorist and officer on same graph), one for velocity (with both motorist and officer on same graph), and one for acceleration (s me as before).

Explanation / Answer

a)
Initial velocity, u = 0
Acceleration, a = 0.525 m/s2
Distance traveled, s = 9.95 m
Take v as the final velocity
Using the formula, v2 - u2 = 2as
v = sqrt [2as] = sqrt[2x 0.525 x 9.95]
= 3.23 m/s.

b)
Using the formula v = u + at
t = (v - u) / a
= (3.23 - 0)/0.525 = 6.16 s

c)
Inital velocity = 3.23 m/s
Final velocity, v = 0
Distance traveled, s = 15.9 m
Using the formula v2 - u2 = 2as
a = - u2/2s = - (3.23)2/(2 x 15.9)
= - 0.328 m/s2

d)
Again using the equation
v2 - u2 = 2as
u = 3.23 m/s, a = - 0.328 m/s2, s = 8.4 m
v2 = u2 + 2as
v = sqrt[u2 + 2as] = sqrt[(3.23)2 + 2 x ( - 0.328) x 8.4]
= 2.22 m/s

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