Answer:
Here, we are required to determine how fast the car is moving, and how far the car goes before the braking period.
A. Vf = acceleration, a1 × time, t1
B. Therefore, D1 = a1 × (t1)².
C. D = {a1 × (t1)² + a2 × (t2)²}.
A. At uniform acceleration, the rate of change of velocity is constant with time.
However, since the car starts from rest, it's initial velocity, Vi = 0.
Therefore, the velocity (speed) at which the car is moving before the braking period, is,
Vf = acceleration, a1 × time, t1
B. To determine how far
the car has moved before the braking period is, D1 = Velocity, Vf × time, t1.
However, velocity, Vf = a1 × t1.
Therefore, D1 = a1 × (t1)².
This is so because the velocity, Vf is the velocity of travel from rest and lasts over time, t1 at which point the braking period commenced.
C. During the braking period, there's deceleration (i.e decay in speed) which returns the velocity of the car back to zero, ultimately bringing the car to a halt.
During the braking period, the total distance covered is also,
D2 = a2 × (t2)².
Therefore, the total distance covered during the motion is, D1 + D2
Total distance, D = D1 + D2.
D = {a1 × (t1)² + a2 × (t2)²}.
Read more:
brainly.com/question/13664094
