A car start from the rest and accelerate at a uniform rate of 5 m/sec ...
Given Information:
- Initial velocity (u) = 0 m/sec
- Acceleration (a) = 5 m/sec2
- Total time (t) = 25 sec
- Average speed (v) = 20 m/sec
Finding:
Time for which the car moves with constant speed.
Solution:
Let's assume that the car moves with constant speed (vc) for time (tc) after accelerating and before decelerating.
Step 1: Finding the distance covered during the accelerated motion
Using the equation of motion:
s = ut + 1/2at2
Where,
- s = Distance covered
- u = Initial velocity = 0 m/sec
- a = Acceleration = 5 m/sec2
- t = Total time taken for the journey = 25 sec
Substituting the values, we get:
s1 = 0 + 1/2 * 5 * (25/2)2 = 1562.5 m
Step 2: Finding the distance covered during the decelerated motion
Using the same equation of motion:
s = vt - 1/2at2
Where,
- s = Distance covered
- v = Final velocity = 0 m/sec
- a = Retardation = -5 m/sec2
- t = Total time taken for the journey = 25 sec
Substituting the values, we get:
s2 = 20 * 25 - 1/2 * (-5) * (25/2)2 = 1562.5 m
Step 3: Finding the distance covered during the motion with constant speed
Using the equation:
vc = s3/tc
Where,
- vc = Constant speed of the car
- s3 = Distance covered during the motion with constant speed
- tc = Time for which the car moves with constant speed
Substituting the values, we get:
s3/tc = 1562.5 m
Step 4: Finding the time for which the car moves with