A car starts from rest and travels for 4.93 s with a uniform acceleration of +1.53 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.99 m/s2. If the brakes are applied for 2.83 s, how fast is the car going at the end of the braking period?
To find the final velocity of the car after the braking period, we need to break down the problem into two parts: the initial acceleration phase and the braking phase.
In the initial acceleration phase:
Given:
Initial velocity (u) = 0 m/s (since the car starts from rest)
Acceleration (a) = +1.53 m/s^2 (positive sign indicates acceleration)
Time (t) = 4.93 s
Using the equation v = u + at, where v is the final velocity, we can calculate the final velocity at the end of the initial acceleration phase:
v = u + at
v = 0 + (1.53)(4.93)
v = 7.55 m/s
So, at the end of the acceleration phase, the car is traveling at a velocity of 7.55 m/s.
In the braking phase:
Given:
Initial velocity (u) = 7.55 m/s (since it continues from the previous phase)
Acceleration (a) = -1.99 m/s^2 (negative sign indicates deceleration)
Time (t) = 2.83 s
Again, using the equation v = u + at, we can calculate the final velocity at the end of the braking phase:
v = u + at
v = 7.55 + (-1.99)(2.83)
v = 2.0 m/s
Therefore, at the end of the braking period, the car is traveling at a velocity of 2.0 m/s.