At the instant a traffic light turns green, a car starts from rest with a given constant acceleration 0.5 m/s^2. just as the light turns green, a bus, traveling with a given cinstant speed 16.0 m/s, passes the car. The car speeds up and passes the bus some time later. How far down the road has the car traveled when the car passes the bus?
Car:
a(t)=0.5
v(t)=0.5t
s(t)=0.25t^2
all of this information is found from taking the integral of 0.5 and since the integral of the acceleration function is v(t)=0.5t+v(initial) we just substitute the initial velocity with 0 because it's 0. Same thing with initial the position function.
bus:
v(t)=16
s(t)=16t
so with this information you're trying to find when s(t) of the car = s(t) of the bus to find when the car catches up to the bus.
0.25t^2=16t
0.25t^2-16t=0
0.25t(t-64)=0
t=0, 64
t=64s
we just ignore that t=0 because time can't equal 0, and most problems will have two solutions (one negative or one equal to 0) and you will be required to cancel one to find the answer.
To find the distance the car has traveled when it passes the bus, we need to determine the time it takes for the car to catch up with the bus.
Let's break down the problem and use the following variables:
- Car's initial velocity (u_car) = 0 m/s (since it starts from rest)
- Car's acceleration (a_car) = 0.5 m/s^2
- Bus's velocity (v_bus) = 16.0 m/s
- Time taken by the car to pass the bus (t)
First, we need to find the time it takes for the car to catch up to the bus. To do this, we can equate the displacement of the car and the displacement of the bus.
The displacement of the car can be calculated using the equation:
S_car = u_car * t + (1/2) * a_car * t^2
Since the car starts from rest, its initial displacement (S_car) can be simplified to:
S_car = (1/2) * a_car * t^2
The displacement of the bus can be calculated as:
S_bus = v_bus * t
Setting the displacement of the car equal to the displacement of the bus, we have:
(1/2) * a_car * t^2 = v_bus * t
To solve for t, we can rearrange the equation:
(1/2) * a_car * t^2 - v_bus * t = 0
Now, we can solve this quadratic equation to find the value of t.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case:
a = (1/2) * a_car = 0.25 m/s^2
b = -v_bus = -16.0 m/s
c = 0
Applying these values, the equation becomes:
t = (-(-16.0) ± √((-16.0)^2 - 4 * 0.25 * 0)) / (2 * 0.25)
Simplifying further:
t = (16.0 ± √(256 - 0)) / 0.5
t = (16.0 ± 16) / 0.5
Using the positive value of t, we find:
t = (16.0 + 16) / 0.5 = 32 / 0.5 = 64 seconds
Now that we have the time it takes for the car to pass the bus, we can calculate the distance traveled by the car using the formula:
S_car = u_car * t + (1/2) * a_car * t^2
Substituting the values we have:
S_car = 0 * 64 + (1/2) * 0.5 * (64^2) = 0 + 16 * 64 = 1024 meters
Therefore, the car has traveled 1024 meters down the road when it passes the bus.