# Kinematics: Motion in One Dimension

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Problem #1.

A car traveling at constant speed of 30m/s passes a trooper hidden behind a bush. Once second after the speeding car passes the bush, the trooper sets off in chase with a constant acceleration of 3.0 m/s2. How long does it take the trooper to overtake the speeding car? How far does the trooper travel during this time?

Problem #3.

A stone is thrown vertically downward with a speed of 6m/s from a cliff. How far down will the stone be from the cliff after 3.0 s? What will its velocity after 3.0s?

Problem #4.

A parachutist with a camera, both descending at a speed of 10m/s, releases that camera at an altitude of 50m. How long does the camera take to reach the ground? What is the velocity of the camera just before it hits the ground?

Problem #5.

A rocket moves upward, starting from rest with an acceleration of +29.4 m/s2 for 4.0s. It runs out of fuel at the end of this 4.0s and continues to move upward. How high does it rise above its original starting point? What is the time of flight of the rocket?

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Problem #1.

A car traveling at constant speed of 30m/s passes a trooper hidden behind a bush. Once second after the speeding car passes the bush, the trooper sets off in chase with a constant acceleration of 3.0 m/s2. How long does it take the trooper to overtake the speeding car? How far does the trooper travel during this time?

Let the trooper takes time t seconds in overtaking at a distance of s m. The initial velocity of the trooper is zero. Then using the second equation of motion we have

s = u*t + ½*a*t2]

s = 0*t + ½ *3*t2

Or s = (3/2) t2 -------------- (1)

Now the time for which the car was in motion before the overtaking is (t + 1) second and in this time it has covered the same distance s at constant velocity and hence

s = v*t = 30*t -------------- (2)

Substituting s from equation (2) in (1) we get

30*t = (3/2) t2

or t2 - 20t = 0

or t(t - 20) = 0

gives t = 0 and t = 20 second, t=0 is corresponding to the initial condition when both are near the bush hence the correct answer is 20 second.

Hence the time taken in overtaking is 20 s.

Substituting in equation (2) we get ...

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