A car starts from rest and travels for 5.0 seconds with a uniform acceleration of +1.5 m/s^2. The driver then applies the brakes, causing a uniform acceleration of -2.0 m/s^2. If the brakes are applied for 3.0 seconds, how fast is the car going at the end of the braking period, and how far has it gone from its start?
There are many ways to do this, some easy, some hard.
Momentum and impulse are conserved.
F*time=final momentum mv
but force=mass*acceleration
m*1.5m/s^2*5sec-m*2.0m/s^2*3sec=m(vf)
solve for Vf
Distance?
distance1=1/2 a t^2=1/2(1.5)5^2
distance2=vi*t-1/2 (2)3^2
where vi=at from the first or vi=1.5*5
so add the distances.
19.0
To solve this problem, we can use the kinematic equations of motion.
First, let's find the final velocity of the car at the end of the first 5.0 seconds when it is accelerating with a uniform acceleration of +1.5 m/s^2. We'll use the equation:
v = u + at
Where:
v is the final velocity
u is the initial velocity (which is 0, as the car starts from rest)
a is the acceleration (+1.5 m/s^2 in this case)
t is the time (5.0 seconds)
Substituting the values into the equation:
v = 0 + (1.5 m/s^2)(5.0 s)
v = 7.5 m/s
So, the velocity of the car at the end of the first 5.0 seconds is 7.5 m/s.
Now, let's find the distance covered by the car during this time. We can use the equation:
s = ut + (1/2)at^2
Where:
s is the distance
u is the initial velocity (0)
t is the time (5.0 seconds)
a is the acceleration (+1.5 m/s^2)
Substituting the values:
s = 0(5.0 s) + (1/2)(1.5 m/s^2)(5.0 s)^2
s = 0 + (1/2)(1.5 m/s^2)(25.0 s^2)
s = (1/2)(1.5)(25.0) m
s = 18.75 m
Hence, the car has traveled a distance of 18.75 meters in the first 5.0 seconds.
Next, let's find the final velocity of the car at the end of the braking period, which lasts for 3.0 seconds with a uniform acceleration of -2.0 m/s^2.
Using the same equation:
v = u + at
Here:
u is the final velocity at the end of the first 5.0 seconds (7.5 m/s)
a is the acceleration (-2.0 m/s^2)
t is the time (3.0 seconds)
Substituting the known values:
v = 7.5 m/s + (-2.0 m/s^2)(3.0 s)
v = 7.5 m/s + (-6.0 m/s)
v = 1.5 m/s (Note: The negative sign indicates that the velocity is in the opposite direction)
So, the velocity of the car at the end of the braking period is 1.5 m/s.
Lastly, let's find the distance covered by the car during the braking period. We can use the same equation as before:
s = ut + (1/2)at^2
Here:
u is the final velocity at the end of the first 5.0 seconds (7.5 m/s)
t is the time (3.0 seconds)
a is the acceleration (-2.0 m/s^2)
Substituting the values:
s = 7.5(3.0 s) + (1/2)(-2.0 m/s^2)(3.0 s)^2
s = 22.5 m + (1/2)(-2.0 m/s^2)(9.0 s^2)
s = 22.5 m + (-9.0 m)
s = 13.5 m
Therefore, the car has traveled a distance of 13.5 meters during the braking period.
In summary, at the end of the braking period, the car is going at a speed of 1.5 m/s and has traveled a total distance of 13.5 meters from its start.