Question

A car acquires a velocity of 56 km/h in 15 seconds
starting from rest.

Find
(0) Acceleration
(ii) Distance travelled in this time.​

Answer 1

Given :-

★ A car acquires a velocity of 56 km/h in 15 sec.

To Find :-

★ Acceleration

★ Distance

Solution :-

According to the question :-

u = 0 m/s

v = 56 km/s

Now, convert the velocity into m/s.

( divide the speed value by 3.6)

Therefore,

v = 56/3.6 m/s

= 15.56m/s

Here, we know

Acceleration is the rate of change of velocity.

$\implies \sf a=\dfrac{v-u}{t}$

On substituting the given values, we get

⟹ a = 1.04 m/s²

Again,

we need to find out the value of S.

$\sf v^2-u^2=2as$

put the values in this equation_

⟹ (15.56)² - (0)² = 2 × 1.04 × s

⟹ s = 116.401 m (Approximately)

Hence,

$\underline{ \overline{ \boxed{ \blue{ \bf{ \: \:a= 1.04m/s² \: cm \: \: }}}}}$

$\underline{ \overline{ \boxed{ \blue{ \bf{ \: \:s= 116.401m \: cm \: \: }}}}}$

Extra Information :-

Velocity ______ v=u+at

Displacement with positive acceleration

___________________s=ut+1/2at²

Displacement with negative acceleration

___________________s=vt−1/2at²

Displacement knowing initial and final speeds

____________________s=1/2(u+v)t