Question

A car starts from rest and moves along the x-axis with constant acceleration

4m s-2 for 10 seconds. If it then continues with constant velocity. what distance

will the car cover in 14 seconds since it started from the rest?​

Answer 1

Answer:

same question here bro ????

Answer 2

Answer:

Since d = vi*t + 0.5 a*t^2 ,

where the final and initial velocity are vf and vi respectly,

Since the car starts from rest, so the initual velocity is equal to zero “vi=0”.

So

At the first interval :

d1= 0 + 0.5 * 5 * (8)^2 = 640 m ,

At the second interval :

The car moves at a constant velocity for another 4 seconds.

we are required to determine the distance at this interval “d2”.

we must determine the final velocity at the first interval so that it could be the final velocity in the second interval.

Since vf = vi + at ,

So vf = 0 + 5*8 = 40 m/sec.

So vi at the second interval"the another 4 seconds” is equal to 40 m/sec.

So d2 = v * t = 40 * 4 = 160 m

So the total distance for 12 seconds from rest = d1+d2 = 640 + 160 = 800 m