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?
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
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.
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.
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
To solve the problem, we can use the equations of motion. Let's break down the problem and solve it step by step.
(i) Find at what distance will the bus overtake the car?
Step 1: Determine the time it takes for the bus to catch up to the car.
We know that both the bus and the car start at the same time. Since the car is traveling with a constant velocity, its acceleration is zero (a = 0). The bus, on the other hand, has a constant acceleration of 5 m/s². To find out when the bus catches up to the car, we need to determine the time it takes.
Using the equation of motion: v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
For the car: v(car) = u(car) + a(car)t = 50 m/s + 0*t (as the car has no acceleration)
For the bus: v(bus) = u(bus) + a(bus)t = 0 + 5*t
Since the bus catches up to the car when their velocities are equal, we can equate these two equations.
50 = 5t
Solving for t, we get t = 10 seconds.
Step 2: Calculate the distance the bus travels in that time.
To find the distance traveled by the bus, we can use the equation of motion: s = ut + 0.5at².
Since the bus starts from rest (u(bus) = 0) and has a constant acceleration of 5 m/s², we can substitute the values to find the distance traveled.
s(bus) = 0*10 + 0.5*5*(10^2) = 0 + 0.5*5*100 = 250 meters.
Therefore, the bus overtakes the car at a distance of 250 meters.
(ii) How fast will the bus be traveling then?
To find the final velocity of the bus when it overtakes the car, we can use the equation of motion: v = u + at.
We know that the initial velocity of the bus (u(bus)) is 0 and the acceleration (a(bus)) is 5 m/s². We can substitute these values into the equation to find the final velocity (v(bus)).
v(bus) = u(bus) + a(bus)t = 0 + 5*10 = 50 m/s.
Therefore, the bus will be traveling at a speed of 50 m/s when it overtakes the car.