Question

At the instant the traffic light goes green, a car waiting at the intersection starts ahead with an acceleration of 3.8m/s^2. At the same instant, a bus coming from behind moving at 25.0m/s passes the car. When will the car catch up with the bus?

Answer 1

This one is pretty straightforward.

You know that at

t=0

x=0

The car starts moving with a constant acceleration of 2.1 m/s2.

At the same time, a truck moving at a speed of 9.2 m/s passes the car.

The idea here is that the distance they travel until the car overtakes the truck is the same for both the car, and the truck.

The same can be said for the time that passes until the two meet again.

So, if the car takes a time t to reach the truck, and both travel a distance d before that happens, then you an say that

d=v⋅t

→ for the truck

and

$d=Vo*t+ </p><p>2</p><p>1$

for car where Vo=0

This will get you

$v*t= </p><p>2</p><p>1$

$*a*t </p><p>2$

$⟹ t= </p><p>a</p><p>2v$

$t= </p><p>2.1 </p><p>s </p><p>2$

m

$2.92 </p><p>s</p><p>m$

= 8.76 s

This means that the distance is

$d=v*t=9.2 </p><p>s</p><p>m$

*8.76=80.6m

The speed of the car is

velocity of car =a⋅t=2.1*8.76s=18.4 m/s