Question

A body starting from rest travels with the uniform acceleration if it travels 100m in 5 sec. What is the distance of acceleration?

Answer 1

Answer:

4ms^-2

Explanation:

acceleration = v-u/t. starting from rest means

u=0

v = s/t

=> 100/5

v = 20m/s

a = v-u/t

=> 20-0/5

a= 4ms^2

Answer 2

Given :-

• Initial velocity,u = 0 m/s
• Distance travelled, s = 100m
• Time taken,t = 5 sec

To Find :-

• The acceleration produced by the body,a

Solution :-

$\small\underline{\pmb{\sf \:Using \: 2nd\: equation \: of \: motion :-}}$

$\: \: \: \: \: \: \: : \implies \underline{ \boxed { \pink{{\frak{s = ut + \frac{1}{2}a {t}^{2} }}}}}$

${ \: \: \: \: \: \: \: : \implies\sf 100 = 0(5)+ \dfrac{1}{2}a {(5)}^{2} }$

${ \: \: \: \: \: \: \: : \implies\sf 100 = 0+ \dfrac{25}{2}a }$

${ \: \: \: \: \: \: \: : \implies\sf 100 = \dfrac{25}{2}a }$

${ \: \: \: \: \: \: \: : \implies\sf 100 \times 2 \: = \: 25a }$

${ \: \: \: \: \: \: \: : \implies\sf \cancel{\dfrac{200}{25} } \: = \: a }$

$\: \: \: \: \: \: \: :\implies \underline{ \boxed{ \sf{ \pink {\pmb{a = \frak{8} \: m \: {s}^{ 2} }}} }}$

$\therefore\:\underline{\textsf{Acceleration of body is \textbf{8m/s².}}}$