How far from the traffic light will the car overtake the truck?

Question

A car starts from a traffic signal with an acceleration which varies with time as a = √t . At the same time, a truck passes the traffic signal at a uniform speed of 15 m/s. How far from the traffic light will the car overtake the truck?

Answer 1

OK, first off, find how far the truck has travelled at time t, which is nice and easy since:

distance = speed * time

s_truck = 15t

Now what about the car? Well, if you integrate the formula for it's acceleration with respect to time, you get the formula for it's velocity, so:

a_car = t^0.5

v_car = (2*t^1.5)/3 + c

And since the car starts from rest:

0 = (2*0^1.5)/3 + c

0 = 0 + c

c = 0

Therefore:

v_car = (2*t^1.5)/3

Next, integrating the formula for velocity with respect to time gets us the formula for displacement, therefore:

v_car = (2*t^1.5)/3

s_car = (4*t^2.5)/15 + d

And since the car starts at 0m from the lights:

0 = (4*0^2.5)/15 + d

0 = 0 + d

d = 0

Therefore:

s_car = (4*t^2.5)/15

Now we have two equations for the displacement of each vehicle, all we need to do is find when they are equal:

s_truck = s_car

15t = (4*t^2.5)/15

225t = 4 * t^2.5

56.25t = t^2.5

56.25 = t^1.5

t = 56.25^(2/3)

t = 3164.0625^(1/3)  <-- cube root of 3164.0625

t = 14.68 seconds

So the truck and the car have travelled the same distance from the lights in 14.68 seconds, and the car overtakes.

How far have they both travelled in that time? We can easily work that out from the formula for the truck:

s_truck = 15t

s_truck = 15 * 14.68

s_truck = 220 m

That's the answer then, 220 metres from the lights.

Edit: Yeah, sorry about that, I made a mistake first time round (actually I made the same mistake twice, once for each integration :P), corrected now.