A car and a motorcycle start from rest at the same time on a straight section of the road. However, the motorcycle starts 29.6 m behind the car. The car accelerates at a uniform rate of 3.9 m/s^2 while the motorcycle accelerates at a constant rate of 4.2 m/s^2. How much time elapses until the motorcycle overtakes the car?

They travel the same time, and 29.6m different in distance.

dm=1/2 am t^2
dc=1/2 ac t^2

subtract the second equation from the first

dm-dc=1/2 (am-ac)t^2
29.6=1/2 (am-ac)t^2
solve for t.

A car traveling at +7.8 m/s accelerates at the

rate of 0.65 m/s
2
for an interval of 1.6 s.
Find vf .
Answer in units of m/s

14.04

To solve this problem, we need to find the time it takes for the motorcycle to overtake the car.

Let's assume that after time 't', the motorcycle overtakes the car.

First, let's find the initial position of both the car and the motorcycle.

For the car, starting from rest, the initial position is 0 meters.

For the motorcycle, starting 29.6 meters behind the car, the initial position is -29.6 meters.

Now, let's find the position of both the car and the motorcycle after time 't'.

The position of the car after time 't' can be calculated using the equation:
Position of car = (Initial Position of car) + (0.5 * Acceleration of car * t^2)

The position of the motorcycle after time 't' can be calculated using the equation:
Position of motorcycle = (Initial Position of motorcycle) + (0.5 * Acceleration of motorcycle * t^2)

Setting the positions of the car and the motorcycle equal, we have:
(0) + (0.5 * 3.9 * t^2) = (-29.6) + (0.5 * 4.2 * t^2)

Simplifying the equation, we get:
1.95 * t^2 = -29.6 + 2.1 * t^2

Rearranging the equation, we have:
0.15 * t^2 = 29.6

Dividing both sides by 0.15, we get:
t^2 = 29.6 / 0.15

Solving for t, we find:
t^2 = 197.333333...

Taking the square root of both sides, we find:
t ≈ 14.03

Therefore, it takes approximately 14.03 seconds for the motorcycle to overtake the car.