Acar moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. its speed as it passes the second point is 15 m/s. (a) calculate the speed at the first point. (b) calculate the acceleration. (c) at what prior distance from the first point was the car at rest?

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

We are going to put our data:

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

If we analyze the situation, we need to have an aceleretarion  greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

For part c:

We are going to use:

b

explanation:

f=ma newton's second law 7x8=56n