Question

a bus is moving with the initial velocity of 6m/s, after applying the breaks its retardation is 0.5m/s^2 and it stopped after 12 find the distance travelled by the bus after applying the breaks​

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Initial velocity (u) = 6ms⁻¹.
• Retardation (a) = -0.5ms⁻² [Retardation is always negative].
• Time (t) = 12 sec.

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• The distance (s).

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

Use the kinematic equations:

$\boxed{\substack{\displaystyle \sf v=u+at \\\\ \displaystyle \sf s=ut+\dfrac{1}{2}at^2 \\\\ \displaystyle \sf 2as=v^2-u^2}}$

It is apt to use the second formula because we are given u, a and t and we need to find s.

$\to \sf{\red{s=ut+\dfrac{1}{2}at^2 }}$

Substituting the values, we get:

$\to \sf s=6 \times 12+\dfrac{1}{2}\times -0.5\times 12^2$

$\to \sf s=72-36$

$\leadsto \sf{\red{ s=36 m.}} \bigstar$

The distance travelled is 36 metres.