A car starts from rest and travels for t1 seconds with a uniform acceleration a1. The driver then applies the brakes, causing a uniform acceleration a2. If the brakes are applied for t2 seconds, determine the following. Answers are in terms of the variables a1, a2, t1, and t2. THIS IS NOT A MULTIPLE-CHOICE QUESTION - I NEED THE EQUATION FOR EACH SECTION

a) How fast is the car going just before the beginning of the braking period?

b) How far does the car go before the driver begins to brake?

c) Using the answers to parts (a) and (b) as the initial velocity and position for the motion of the car during braking, what total distance does the car travel?

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}.

Explanation:

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 last 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}.

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a and b sometimes have a positive value