Solve this : Mods And StarsA car starts from rest and moves along the

1.	Solve this : Mods And StarsA car starts from rest and moves along the x-axis with constant acceleration of 5 m/s² for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?
Answer» Given: A CAR starts from rest and moves along the x-axis with a constant ACCELERATION of 5 m/s² for 8 seconds. To Find: what distance will the car cover in 12 seconds since it started from rest. Solution: We know that, $s = ut + \frac{1}{2}at^{2}$ But given that, the initial velocity is zero. Therefore, the distance covered by the car in first 8 seconds is $s = 0 + \frac{1}{5} \times 5 \times {8}^{2}$ s = 180 metres Therefore, the velocity ATTAINED at the end of 8 seconds is: from, v = u + at =>V= 0 + 5 x 8 40 m/s Therefore, since the car has moved with constant velocity from 8th second to 12 second, the distance travelled is S = v × t = 40 x 4 = 160 m. Therefore, the total distance travelled is 160 + 160 = 320m. Hope it helps :)

Solve this : Mods And StarsA car starts from rest and moves along the x-axis with constant acceleration of 5 m/s² for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

Answer»

A CAR starts from rest and moves along the x-axis with a constant ACCELERATION of 5 m/s² for 8 seconds.

what distance will the car cover in 12 seconds since it started from rest.

We know that,

$s = ut + \frac{1}{2}at^{2}$

But given that, the initial velocity is zero. Therefore, the distance covered by the car in first 8 seconds is

$s = 0 + \frac{1}{5} \times 5 \times {8}^{2}$

s = 180 metres

Therefore, the velocity ATTAINED at the end of 8 seconds is:

from, v = u + at

=>V= 0 + 5 x 8 40 m/s

Therefore, since the car has moved with constant velocity from 8th second to 12 second, the distance travelled is