A car moving with a constant acceleration covers the distance between two points 80 m apart in 8.0 s. Its velocity as it passes the second point is 14 m/s.
(a) What was the speed at the first point?
1 m/s
(b) What is the constant acceleration?
2 m/s2
(c) How far behind the first point was the car at rest?
3 m
To answer these questions, we need to use the equations of motion that relate distance, time, velocity, and acceleration. The main equation we will be using is:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
(a) To find the speed at the first point (u), we know the final velocity (v) and the time (t). Rearranging the equation, we have:
u = v - at
Substituting the given values:
v = 14 m/s
t = 8.0 s
u = 14 m/s - a * 8.0 s
Since the acceleration is constant, we know a = change in velocity / change in time. Therefore:
a = (v - u) / t
Substituting the given values:
a = (14 m/s - u) / 8.0 s
Now we have a system of equations. We can solve for both u and a.
(b) To find the constant acceleration, we rearrange the equation:
a = (v - u) / t
Substituting the given values:
a = (14 m/s - u) / 8.0 s
Solving for a, we can use the acceleration in part (b) to find u.
(c) To find how far behind the first point the car was at rest (d), we need to determine the distance traveled by the car from the first point to the second point. We can use the equation of motion:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity
t = time
a = acceleration
Since the car passes the second point with a final velocity of 14 m/s, we can use this equation to find the distance traveled:
80 m = ut + (1/2)at^2
Using the values of u and a obtained from parts (a) and (b), we can solve for t. Once we have t, we can substitute it back into the equation to find d.