Ask a New Question

Asked by ronit

a car starting from rest acquired a speed of 25ms-¹ in 10sec,after which it maintain this speed for 10sec. find (a)the acceleration (b) distance traveled during acceleration (c)total distance traveled?
10 years ago

Answers

Answered by Steve
acceleration is (25m/s)/(10s) = 2.5 m/s^2
distance during acceleration: s = 1/2 at^2 = 125m
so, total distance is 125 + 10*25 = 375 m
10 years ago

Related Questions

a car starting from rest is accelerated at 2m/s(square) for 10s, then moves at constant velocity for... Starting from rest at 10m to the left of the origin (the starting line of a race), a go-kart undergo... A car starting from rest is uniformly accelerated so that it's velocity in 5 seconds is 36 km/h . It... a car starting from rest is accelerated to uniformly to a velocity of 1oo.2km/h,a break is applied i... A car starting from rest is accelerated uniformly to a velocity of 100.8km/h.A Break is applied in 4... Starting from rest, a car undergoes a constant acceleration of 6 m/s2. What is the final velocity af... A car starting from rest at the top of a hill at point A and coasting with friction to the botton po... A car starting From rest is accelerated to a velocity of 28m/s, a brake is apply in 4 second. Calcul... A car starting from rest is uniform accelerated,so that its velocity is 6sec in 58m/h a brake is the... a car, starting from rest , began moving in straight line and its velocity is given by v= (8t2+6t)...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use