A bus starts from rest with a constant acceleration of 5m/s². At the same time a car traveling with a constant velocity of 50m/s overtakes and passes the bus. (i) Find at what distance will the bus overtakes the car? (ii) How fast will the bus be travelling then?

Answer 1

The question is poorly worded.
It says, "At the same time" a car .... overtakes and passes the bus. So the time would be zero.

I will assume that the car was a certain distance behind the bus

bus:
a = 5
v = 5t + c
but when t=0, v=0 , so c=0
v = 5t
s = distance = (5/2)t^2 + k
again, when t = 0 , s = 0 , so k = 0

s = 2.5t^2

in t seconds the car went 50t m

so 2.5t^2 = 50t
t^2 = 20t
t^2 - 20t = 0
t(t-20) = 0
t = 0 , the trivial solution
or
t = 20

when t = 20 s, both the car and the bus went :
s = 50(20) m = 1000 m or 1 km
at that time, the bus went at 5t or 100 m/s

check:
at 50 m/sec , in 20 seconds the car went 1000 m
in 20 seconds, the bus went 2.5t^2 or 2.5(20^2) = 1000 m

Answer 2

Even though the above solution is mathematically correct, it makes no sense and the question is ridiculous.
let's look at what time their speeds are the same

5t = 50
t = 10
after 10 seconds, the bus is going at 50 m/s
and the car is going at a constant speed of 50m/s
but the car has not passed the bus yet, and the bus is still "speeding up".
So the car can never pass the car.

btw,
50 m/s = .05 km/(1/3600 hr) = 180 km/h , a rather dangerous speed for a car to be passing a bus.

Answer 3

Eh? I read it that the car passes the bus just as the bus starts to move. The car has constant speed, so at some point the accelerating bus will overtake the car.

I haven't checked the math above, but the problem makes sense to me.

Answer 4

Steve is right, I misread the question

Nevertheless, that is some dangerous speed the bus is going
The math is correct, the bus will catch the car after 20 seconds