Question

A speeder passes a parked police car at a constant speed of 25.1 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.56 m/s 2 . How much time passes before the speeder is overtaken by the police call?

Answer 1

Answer:

The police car needs 19.6 s to catch the speeder.

Explanation:

Assuming a straight motion, these are the kinematic equations that we could use to solve the problem:

For the speeder that moves at a constant speed, the equation to determine its position after a time "t" is:

X = X0 + v*t

For the police car, which has a uniform acceleration, the position after a time "t" is:

X = X0 + v0*t + 1/2*a*t²

where:

X = position

X0 = initial position

v = velocity

v0 = initial velocity

t = time

a = acceleration

We have to find how much time needs the police car to catch the speeder, that means, we have to find at which time the position of both cars is the same. So we have to equalize both equations and obtain "t".

Position of speeder = position of the police car

X0 + v*t = X0 + v0*t +1/2*a*t²

Both cars have the same initial postion (X0) and the police car starts from rest (v0 = 0). Then:

v*t = 1/2*a*t² (multiplying both sides by 2)

2v*t = a * t² (dividing both sides by a*t)

2v/a = t

Now, we replace with the data that give us the problem:

2(25.1 m/s) / 2.56 m/s² = t

19.6 s = t

Answer 2

Answer:

19.61 second

Explanation:

speed of speeder car, u = 25.1 m/s

initial speed of police car, u = 0

acceleration of police car, a = 2.56 m/s^2

let it take time t to overtake the speeder car.

For overtaking, the distance traveled by the speeder car in time t is same as the distance traveled by the police car.

Distance traveled by the speeder car, d = speed x time

d = 25.1 x t ..... (1)

Distance traveled by police car is given by second equation of motion

$d = ut + 1/2 at^{2}$

$d= 0 \times t + 1/2 \times 2.56\times t^{2}$

d = 1.28t^2     .... (2)

From equation (1) and (2) we get

25.1 t = 1.28 t^2

25.1 = 1.28 t

t = 19.61 second