Question

A scooter moving at a speed of 10m/s is stopped by applying breakes which produce a uniform acceleration of, -0.5 m/s². How much distance will be covered by the scooter before it stops.​

Answer 1

Solution -

We have,

• Initial speed (u) = 10m/s
• Final speed (v) = 0
• Acceleration (a) = -0.5m/s²

We have to calculate the distance covered (s) by the scooter before it stops.

⠀

Using third equation of motion

$\large{\bf{\longmapsto{\boxed{\pink{v^2 = u^2 + 2as}}}}}$

Putting values

$\tt\dashrightarrow{(0)^2 = (10)^2 + 2 \times (-0.5) \times s}$

$\tt\dashrightarrow{0 = 100 - 1 \times s}$

$\tt\dashrightarrow{0 = 100 - s}$

$\bf\dashrightarrow{s = 100\: m}$

$\: \underline{\sf{Thus,\: the\: distance\: covered\: is\: 100\: metres.}}$

Answer 2

question :-

A scooter moving at a speed of 10m/s is stopped by applying breakes which produce a uniform acceleration of, -0.5 m/s². How much distance will be covered by the scooter before it stops

Explanation:

u = 12 m/s

v = 0 m/s

a = - 0.6 m/s^2

S = ???

$s = \frac{ {v}^{2} - {u}^{2} }{2a}$

$= \frac{0 - 144}{ - 2 \times 0.6}$

$= 120 \: m$

Therefore, the scooter stops after 120 m.