A car moving with constant acceleration covers the distance between two points 60 m apart in 6 seconds.Its speed point is 15 m/s.At what prior distance from the first point was the car at rest?
To find the distance from the first point where the car was at rest, we can use the equation of motion:
\[d = ut + \frac{1}{2}at^2\]
Where:
d = distance traveled
u = initial velocity
t = time
a = acceleration
Given that the car covers a distance of 60 m in 6 seconds with a speed of 15 m/s, we can determine the acceleration using the equation:
\[v = u + at\]
Substituting the known values, we get:
15 m/s = u + a * 6 s
Since the car started from rest, its initial velocity, u, is 0 m/s. Thus, the equation becomes:
15 m/s = 0 m/s + a * 6 s
Simplifying the equation, we find:
15 m/s = 6 a
Dividing both sides by 6, we get:
a = 15 m/s / 6 s
a ≈ 2.5 m/s^2
Now, using the equation of motion, we can find the distance from the first point where the car was at rest:
0 = 0 * t + (1/2) * 2.5 m/s^2 * t^2
Simplifying the equation, we get:
0 = 1.25 m/s^2 * t^2
Since the distance from the first point is 60 m, we can solve for t:
60 m = 1.25 m/s^2 * t^2
Dividing both sides by 1.25 m/s^2, we get:
48 s^2 = t^2
Taking the square root of both sides, we find:
t ≈ √48 s
t ≈ 6.928 s
Therefore, the car was at rest approximately 6.928 seconds before reaching the first point.