A car moving with constant acceleration covered the distance between two points `60.0 m` apart in `6.00 s.` Its speed as it passes the second point was `15.0 m//s.`
(a) What is the speed at the first point?
(b) What is the acceleration?
(c) At what prior distance from the first was the car at rest?

Explanation

To solve the problem step by step, we will break it down into three parts as per the questions asked.

### Given Data:
- Distance between two points, \( S = 60.0 \, \text{m} \)
- Time taken, \( t = 6.00 \, \text{s} \)
- Final speed at the second point, \( v = 15.0 \, \text{m/s} \)

### (a) Finding the speed at the first point (initial speed \( u \)):
We can use the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
However, we need to first find the acceleration \( a \). We can use another equation:
\[ v = u + at \]
We will need to express \( a \) in terms of \( u \) and \( v \).

From the equation \( S = ut + \frac{1}{2} a t^2 \), we can rearrange it to find \( a \):
\[ a = \frac{2(S - ut)}{t^2} \]
Substituting \( S = 60.0 \, \text{m} \) and \( t = 6.0 \, \text{s} \):
\[ a = \frac{2(60 - 6u)}{36} \]
\[ a = \frac{120 - 12u}{36} \]
\[ a = \frac{10 - u}{3} \]

Now substituting \( a \) into the equation \( v = u + at \):
\[ 15 = u + \left(\frac{10 - u}{3}\right) \cdot 6 \]
\[ 15 = u + 2(10 - u) \]
\[ 15 = u + 20 - 2u \]
\[ 15 = 20 - u \]
\[ u = 20 - 15 = 5 \, \text{m/s} \]

### (b) Finding the acceleration \( a \):
Now that we have \( u = 5 \, \text{m/s} \), we can substitute \( u \) back into the equation for \( a \):
\[ a = \frac{10 - u}{3} = \frac{10 - 5}{3} = \frac{5}{3} \, \text{m/s}^2 \]

### (c) Finding the prior distance from the first point where the car was at rest:
To find the distance \( s \) from the point where the car was at rest, we can use the equation:
\[ v^2 = u^2 + 2as \]
Here, \( v = 5 \, \text{m/s} \) (the speed at point A), \( u = 0 \) (the speed when the car was at rest), and \( a = \frac{5}{3} \, \text{m/s}^2 \):
\[ (5)^2 = 0 + 2 \left(\frac{5}{3}\right) s \]
\[ 25 = \frac{10}{3} s \]
\[ s = \frac{25 \cdot 3}{10} = \frac{75}{10} = 7.5 \, \text{m} \]

### Final Answers:
(a) Speed at the first point: \( 5 \, \text{m/s} \)
(b) Acceleration: \( \frac{5}{3} \, \text{m/s}^2 \)
(c) Distance from the first point where the car was at rest: \( 7.5 \, \text{m} \)