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?

Question

anonymous · Answer

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

anonymous · Answer

$$(1)D(t)=1/2.α.t^{2}+β.t+C$$
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.

anonymous · Answer

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

anonymous · Answer

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