A car starts from rest and travels for 5.9 s with a uniform acceleration of +3.0 m/s2. The driver then applies the brakes, causing a uniform deceleration of 1.3 m/s2.
If the brakes are applied for 2.1 s, how fast is the car going at the end of the braking period?
Answer in units of m/s
To determine the final velocity of the car at the end of the braking period, we can use the kinematic equations. The relevant equation in this case is:
v = u + at
where v is the final velocity of the car, u is the initial velocity of the car, a is the acceleration, and t is the time.
Since the car starts from rest, the initial velocity (u) is 0 m/s. The car travels with a uniform acceleration of +3.0 m/s² for 5.9 seconds. Using the equation above, we can find the velocity of the car after this period:
v1 = 0 + (3.0 m/s²)(5.9 s)
v1 = 17.7 m/s
After 5.9 seconds, the driver applies the brakes, causing a uniform deceleration of -1.3 m/s². The negative sign indicates deceleration, or the opposite of acceleration. The braking period lasts 2.1 seconds.
Using the same equation, we can find the velocity of the car at the end of the braking period:
v2 = v1 + (-1.3 m/s²)(2.1 s)
v2 = 17.7 m/s + (-1.3 m/s²)(2.1 s)
v2 = 17.7 m/s - 2.73 m/s
v2 = 14.97 m/s
Therefore, the car is traveling at a speed of approximately 14.97 m/s at the end of the braking period.