Question

. A car initially at rest starts moving with a constant acceleration of
0.5 ms-2 and travels a distance of 36 m. Find its final velocity.

Answer 1

Given :

• Car is initially at rest [u =0]

• Acceleration of the body is 0.5 m/s² [a = 0.5 m/s²]

• Distance travelled is 49m [s = 49m]

To find :

• Final velocity of the body

• Time taken by the body

SOLUTION :

Firstly lets calculate the time taken by the body through the second equations of motion

$\underline{\boxed{\frak{\pm{\green{\ \: s = ut + \dfrac{1}{2} at {}^{2}}}}}}\:\red{\bigstar}$

✞ s = 49

✞ u = 0

✞ a = 0.5

Substituting the values,

$:\implies \sf \: 49 = 0(t) + \dfrac{1}{2} (0.5)(t {}^{2} )$

$:\implies \sf \: 49 = 0 + \dfrac{1}{2} ( \dfrac{1}{2} )(t {}^{2} )$

$:\implies \sf \: 49 = 0 + \dfrac{1}{4}(t {}^{2} )$

$:\implies \sf \: 49 = \dfrac{t {}^{2} }{4}$

$:\implies \sf \: 49 \times 4= t {}^{2}$

$:\implies \sf \: t = \sqrt{49 \times 4}$

$:\implies \sf \: t = \sqrt{49 } \times \sqrt{4}$

$:\implies \sf \: t = 7 \times 2$

$:\implies\bf \: t = 14s$

So, time taken by the body is 14 s

From,

First equations of motion we find the final velocity of the body :-

$\underline{\boxed{\frak{\pm{\green{\ \: v = u + at}}}}}\:\red{\bigstar}$

✞ u = 0

✞ a = 0.5 m/s²

✞ t = 14 s

$\sf:\implies \: v = 0 + (0.5)(14)$

$\sf:\implies \: v = 0 + ( \dfrac{1}{2} )(14)$

$\sf:\implies \: v = 0 + 7$

$\bf :\implies\: v = 7m/s$

So , final velocity of the body is 7 m/s