Question

a car accelerates from rest to 6.1 m/s in 40 seconds. then continues at this velocity for 60 seconds. then decelerates to a stop in 10 seconds if car traveled in straight line determine final distance from starting point

Answer 1

Break the motion into three phases.

1) Accelerating from rest to 6.1 m/s in 40 s:
- s1 = 0.5 * v * t = 0.5 * 6.1 * 40 = 122.0 m

2) Constant speed 6.1 m/s for 60 s:
- s2 = v * t = 6.1 * 60 = 366.0 m

3) Decelerating to stop in 10 s:
- average speed = (6.1 + 0)/2 = 3.05 m/s
- s3 = average speed * t = 3.05 * 10 = 30.5 m

Total distance = s1 + s2 + s3 = 122.0 + 366.0 + 30.5 = 518.5 m

So the car is 518.5 m (≈ 519 m) from the starting point.