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

Answer:

84m

Explanation:

Given,

a=2.5m/s^2

u=16m/s

t=4s

Using 2nd Equation of Motion

s=ut+1/2at^2

s=16×4+1/2×2.5×4^2

s=64+1/2×2.5×16

s=64+2.5×8

s=64+20

s=84m

Answer 2

Distance covered = 44 m

Given

A car accelerates uniformly with 2.5 m/s² for 4 seconds and attains a velocity of  16 m/s

To Find

Distance covered by the car

Knowledge Required

$\bf \blue{\bigstar\ \; v=u+at}$

$\bf \red{\bigstar\ \; s=ut+\dfrac{1}{2}at^2}$

where ,

• s denotes displacement
• u denotes initial velocity
• t denotes time
• a denotes acceleration
• v denotes final velocity

Solution

• a = 2.5 m/s²
• t = 4 s
• v = 16 m/s

Apply 1st equation of motion ,

$\bf v=u+at\\\\\to \rm 16=u+(2.5)4\\\\\to \rm 16=u+10\\\\\to \bf u=6\ m/s$

Apply 2nd equation of motion ,

$\bf s=ut+\dfrac{1}{2}at^2\\\\\to \rm s=(6)(4)+\dfrac{1}{2}(2.5)(4)^2\\\\\to \rm s=24+20\\\\\to \bf \green{s=44\ m\ \; \bigstar}$

So , Distance covered = 44 m