Question

A speeder is traveling at 75 mi/h. He passes a standing police car, which starts to chase him. A speeder is traveling at 75 mi/h. He passes a standing police car, which starts to chase him. The police car accelerates from 0mi/h to 85 mi/h in 13 s and travels at 85 mi/h there after.
a. How far from its starting point does the police car overtake the speeder?
b. What is the elapsed time?

Answer 1

the distance traveled by the speeder and police car is same at the instant when police car overtakes the speeder. the distance traveled by speeder is s=(75/3600)*t when t is in secs. the distance traveled by speeder=distance during acc.+distance after acc. distace durig acc=ut+(at^2)/2 a=85/(13*3600) distance after acc.=(85/3600)*(t-13) distance travelled by speeder=speed before acc.+speed after acc. after calculation t=55.25secs substitute t=55.25 in s=(75/3600)*t thus s=1.151mi

Answer 2

\[(1)D(t)=1/2.α.t^{2}+β.t+C\]
Then, you need to divide your problem into two parts.
1/ the police car accelerates
2/ the police car doesn't accelerate but has a constant speed.

The first part is the much complex. Indeed, you don't know yet the acceleration, but it's not really a problem because you can calculate it :
\[(2)V(t)=α.t+β\]
and you know "t" and V(t), also β=0 (the police car starts off).
You also calculate the traveled distance during this 13 secondes thanks to (1).

The second part, you have just to change the initial condition : C and β are not null, but α is null (the car doesn't accelerate you remember). C is the distance you just have calculated, and β is speed (85 mi/h).

I let you search a little the following. You will solve an egality.

Answer 3

I too got adithya252's answer......I think he is right....

Answer 4

After the police car acceleration, the equations are :
\[D_{Police Car}(t) = (85/3600).(t-13) + D_1 \]
\[D_1 = 1/2*85/(13*3600).(13)²\]

\[D_{Speeder}(t)=(75/3600).(t)\]
So, to resolve the egality :
\[D_{police Car}(t)=D_{Speeder}(t)\]
\[ (85/3600).(t-13) + D_1 = (75/3600).(t)\]
\[t=13*85/20\]

I find the same result. t= 55,25 s and d=1,151 mi