Question

A car accelerates uniformly with 2.5ms for 4 seconds and attains a velocity of 16ms-1. Calculate the distance covered
by car during this period.​

Answer 1

Given :

accerlation of the car, a = 2.5 m/s²

time taken, t = 4 s

final Velocity of the car, v = 16 m/s

To Find :

Distance covered by the car

Solution :

As we are provided with final velocity, time taken and accerlation we should use equation of motion.

first of all we have to find final velocity

so, using 1st equation of motion,

${ \dashrightarrow{\bf {v=u+at}}}\\\\{\dashrightarrow {\sf {16=u+(2.5)4}}}\\\\{\dashrightarrow {\sf {16=u+10}}}\\\\{\dashrightarrow{\bf{ u=6\ m/s}}}$

final velocity = 6m/s.

now, applying 2nd equation of motion .i.e.

${ \dashrightarrow{\bf {s=ut+\dfrac{1}{2}at^2}}}\\\\{\dashrightarrow{\sf {s=(6)(4)+\dfrac{1}{2}(2.5)(4)^2}}}\\\\{\dashrightarrow {\sf {s=24+20}}}\\\\{\dashrightarrow{ \boxed {\bf {s=44\ m}}}}$

thus, the Distance covered by the car is 44 m.