Question

Assume, unless otherwise specified, that all numbers have at least 3 significant figures. and please circle your answers. You may work together, but make sure that you are working-not just watching. Peter Policeman is sitting by the side of the road when Sammy Speeder flies by in his Tesla Model S at 100 mph. Peter takes one second to react before jamming on his accelerator giving his Ford Crown Victoria Police Interceptor an acceleration of 3.5 m/s^2. Assume the car can maintain this acceleration up to a maximum speed of 130 mph. How far away from where Peter is sitting does he catch Sammy? (Yes Sammy could go faster, but let's not get carried away here) How long does it take Peter to overtake Sammy, measured from when Sammy passes Peter? Sketch a one graph of the position of each as a function of time.

Answer 1

if t is the time taken to catch sammy to be caught then the distance traveled is

$d= V_s t$

Time

$t =\frac{d}{V_s} = \frac{d}{100 (1000/3600)} = d/27.78 s$ .....(1)

Now the policeman catches Sammy in the same time. Police car take some time to accelerate to maximum speed and then continue on it. Distance travelled to reach the maximum speed is

$V_f^2 - V_i^2 = 2 a d_1$

$\left ( 135 (1000/3600) \right ) - 0 = 2 (3.5) d_1$

$d_1 =200.9m$

Time taken to reach the top speed is

$t_1 = V_f / a = \frac{130 (1000/3600)}{3.5} = 10.32s$

After attaining the top speed policeman has to travel certain distance to catch Sammy

$d_2 = \left ( 130 (1000)/3600 \right ) t_2$

$t_2 = d_2 / \left ( 130 (1000)/3600 \right ) =\left ( d- 200.9 \right ) / 36.11$

Total time taken is

$t = t_1 + t_2 = 10. 32 + \left ( d- 200.9 \right ) / 36.11$ .....(2)

Equating (1) & (2)

$d/ 27.78 = 10. 32 + \left ( d- 200.9 \right ) / 36.11$

Solving for d

d = 572.8 m

b)

Time taken to catch sammy is

$t = 572.8/27.78 = 20.62s$

c)