A car starts from rest and travels for 6.0 s with a uniform acceleration of +1.5 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.0 m/s2. If the brakes are applied for 3.0 s, how fast is the car going at the end of the braking period?
To solve this problem, we can break it down into two parts: the first part where the car is accelerating, and the second part where the car is decelerating.
In the first part, the car starts from rest and travels for 6.0 s with a uniform acceleration of +1.5 m/s^2. We can use the kinematic equation to find the final velocity (v) during this period:
v = u + at
where:
v is the final velocity
u is the initial velocity (which is 0 m/s since the car starts from rest)
a is the acceleration (+1.5 m/s^2)
t is the time (6.0 s)
Plugging in the values, we get:
v = 0 + (1.5)(6.0)
v = 9.0 m/s
So, at the end of the acceleration period, the car is traveling at 9.0 m/s.
In the second part, the driver applies the brakes, causing a uniform acceleration of -2.0 m/s^2. We can use the same kinematic equation to find the final velocity during this period:
v = u + at
where:
v is the final velocity
u is the initial velocity (which is 9.0 m/s since the car is already moving)
a is the acceleration (-2.0 m/s^2)
t is the time (3.0 s)
Plugging in the values, we get:
v = 9.0 + (-2.0)(3.0)
v = 9.0 - 6.0
v = 3.0 m/s
So, at the end of the braking period, the car is traveling at 3.0 m/s.