Question

A car moves along an x axis through a distance of 950 m, starting at rest (at x = 0) and ending at rest (at x = 950m). Through the first 1/4 of that distance, its acceleration is +3.60m/s2. Through the next 3/4 of that distance, its acceleration is -1.20 m/s2 . What are its travel time through the 950 m and its maximum speed

Answer 1

Given, the car starts from rest, so its initial velocity u = 0

Acceleration, (a) = 5 ms-2 and time (t) = 8 s

From first equation of motion,

v = u + at

On putting a = 5 ms-2and t = 8 s in above equation, we get

v = 0+ 5×8= 40ms’1

So, final velocity v is 40 ms’1.

Again, from second equation of motion,

So, the distance covered in 8 s is 160 m. -­

Given, total time t = 12 s.

After 8 s, the car continues with constant velocity i.e., the car will move with a velocity of 40 ms-1.

So, remaining time t’= 12 s-8 s= 4s

The distance covered in the last 4s (s’) =Velocity x Time [∴ Distance = Velocity x Time]

= 40 x 4 = 160 m

[We have used the direct formula because after 8 s, car is moving with constant velocity i.e., zero acceleration].

Total distance travelled in 12 s from the start

D = s + s’= 160+ 160= 320 m

hope it helps you