Simple physics problem, what am I doing wrong?

Question

A police car traveling a constant 93.0km/hr (25.83 m/s), is passed by a speeder traveling 142km/hr (39.44 m/s). Precisely 1.22s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 1.87m/s2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at constant speed)?


The police needs to reach the same speed as the other car, so the final velocity is 39.44 m/s. The initial velocity is 25.83 m/s. Acceleration is 1.87 m/s^2 and the time which we are looking for would be +1.22 seconds because that's how long he took to react. So I plugged it into Vf = Vi + a(t+1.22) which for me turned out to be 8.50 s. I even tried subtracting the time t-1.22s and still got the problem wrong. Any hints?

Answer 1

Ok, where you are going wrong is the first sentence after you re-state the question. If the police reaches the same speed as the other car, then the distance between them will be constant, not zero. (Put yourself in the police car. If a car is ahead of you and you match its speed, the car will stay the same distance in front of you.) The police car needs to go faster than the other car in order to catch up.

Here's a way to think about it. From the point where the speeder passed the cop to where the cop ultimately catches the speeder, the distance is the same for both automobiles. So, you get an expression for the distance the speeder travels, (This is easy since he's moving at a constant speed) 39.44t. Then, get an expression for the distance the cop travels. This has to be done in two parts: before the cop hits the accelerator and after.
First part: 25.83*1.22 = 31.5126. Second part: use the formula - x = v0*t + .5*a*t^2. So, it's 25.83(t-1.22)+ .5*1.87(t-1.22)^2. Hence the total expression for the distance the cop travels is:
31.5126 + 25.83(t-1.22) + .5*1.87(t-1.22)^2. Set this equal to 39.44t (the distance the speeder traveled) and solve.

I came up with 16.91s. Hope that helps.

Answer 2

the problem is in your assumption that the final velocities are the same, they are not. The police car has to go faster than the speeder in order to catch up. Their final positions are the same. Here is how you solve it:

position of the speeder:

x= speeders velocity * (t + 1.22)

this puts the speeders position in terms of time. The 1.22 represents the lag time of the cop.

Position of the police car:

x= velocity of the police car initially *t + 1/2 acceleration *(t squared)

set both x's equal and solve for t.

Answer 3

After reading your question carefully my previous answer does not answer the actual question you want the answer to.

you do not want to know how long it takes the police car to reach 39.44 m/s, you want to know how long it takes for the two cars to be at the same distance downrange.

So, at the moment the car passes traveling at 39.44m/s 1.22 sec goes by. So the car traveled 48.1 m in that time. In that same time the police car, traveling at 25.83 m/s, travels 31.5 m.

so the police car is -16.6 m behind the speeding car.

this is the configuration i will use for t = 0.

now at the moment of interception the speeding car will have traveled:

d_int = (39.44)*t_int

and the police car:

d_int = -16.6 + (25.83)*t_int + (1/2)(1.87)t_int^2

set these equal and solve:

(39.44)*t_int = -16.6 + (25.83)*t_int + (1/2)(1.87)t_int^2

you get:

t_int = 15.69
or
t_int = -1.13 (can't be this one on physical basis)

so
t_int = 15.69 seconds.

Lastly, the question wanted to know "How much time elapses AFTER the police car was passed?" So add 1.22 sec, because i used t = 0 to get the above answer:

After the police car was passed, 16.91 seconds elapsed before the police car reaches the speeding car.

**Also: s_m_hc got it correct first, so s_m_hc deserves the credit. Micheal H was close, but forgot to take into account the distance the police car travels in the 1.22seconds.

Answer 4

The final velocity of the policeman is more than 39.44 m/s. Remember the police must eventually move faster than the speeder in order to catch up.

Answer 5

initially S---------------->25.83 m/s
P---------------->39.44 m/s

Finally S---------------------------------------------S
P---------------------------------------------P
<------------------s----------------------->

let t be the time after police car catches the speeder.
So the distance s, the police car takes in this time t is

s=u.t+(1/2).a.t^2

the speeder will also take the same time t while going to the same distance s. So
s=v(39.44 m/s).t

hence
s(police man) = s(speeder)
u.t+(1/2).a.t^2 = v.t
25.83t+(1/2).1.87.t^2 = 39.44t

while solving this quadratic equation we get two values of t
as t=0 and t=14.556s

So the correct answer is 14.556 seconds.

Answer 6

Is the 352 the height above the ground or the height above the top of the building?