# Police car's acceleration physics problem

A police car traveling a constant 95 km/h is passed by a speeder traveling 140 km/h. Precisely 1.00 s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.00 m/s2, how much time passes after the speeder passes before the police car overtakes the speeder (assumed moving at constant speed)?

I converted 95 km/h to 40 m/s, and 95 to 26 m/s, then I use the equation

x = x + v*t + 1/2 * a *t^2
40 m/s * t + 40 m = 0 + 26 m/s * t +1/2*2.0*t^2

t=15, but that's wrong....any ideas?

Tide
Homework Helper
$$95t+\frac{1}{2}at^2=140(t+1)$$

... where a is the acceleration with appropriate units, of course! :-)

I still get t=15 with that equation.

Tide
Homework Helper
You did? I got 6.3 minutes if I did my arithmetic right - seems too long though! Check to see if you converted the units on your acceleration correctly.

well, since the final answer is in seconds, I converted 140 km/h and 95km/h to m/s. So

95km/h * 1h/60min * 1min/60s * 1000m/1km= 26.389 m/s and
140km/h * 1h/60min * 1min/60s * 1000m/1km= 38.889 m/s.

so 26.389*t+1/2*2.0*t^2=38.889(t+1)
and t=15 s.

Tide
Homework Helper
Oh! The t+1 in my original equation was incorrect. I was mixing seconds with hours. When I correct for that I get something closer to what you have (15.9s).

The hw is online and when I entered 15s and submitted, it said it was incorrect.

Chronos
Gold Member
You are getting distance, not acceleration, from that formula. The original formula Tide gave is correct. You need only solve for time. Keep trying.

Last edited:
Chronos, can you clarify what you said?

Doc Al
Mentor
Tide said:
$$95t+\frac{1}{2}at^2=140(t+1)$$
This isn't quite right for several reasons.

Measuring everything from the moment the speeder passes the police car (at t = 0):
Distance police car travels: $x_p = v_p t + \frac{1}{2}a(t-1)^2$, note that acceleration only begins at t = 1 second, $v_p$ is the initial speed of the police car in m/s.
Distance speeder travels: $x_s = v_s t$, $v_s$ is the speed of the speeder in m/s.

So, when the police car overtakes the speeder:
$$v_p t + \frac{1}{2}a(t-1)^2 = v_s t$$
All times measured in seconds. Now plug in your values and solve for t.

Tide