Question

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

Answer 1

Given

A scooter moving at a speed of 10m/s is stopped by applying brakes which produce a uniform Acceleration of -0.5 m/s² .

To find

How much distance will be covered by the scooter before it stops ?

Solution

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

According to third equation of motion

→ v² = u² + 2as

→ (0)² = (10)² + 2 × (-0.5) × s

→ 0 = 100 - s

→ s = 100 m

Hence, distance travelled by scooter is 100m

Note : (-) minus shows retardation

Answer 2

GiVeN :-

A scooter moving at a speed of 10 m/s is stopped by applying brakes which produce a uniform Acceleration of -0.5 m/s².

• Initial velocity (u) = 10 m/s
• Final velocity (v) = 0 m/s
• Acceleration, or retardation (a) = -0.5 m/s²

To DeTeRmInE :-

The distance covered (s) covered by the scooter before it stops.

AcKnOwLeDgEmEnT :-

$\leadsto$ The first equation of motion :-

$\boxed{\tt{v=u+at}}$

$\leadsto$ The second equation of motion :-

$\boxed{\tt{s=ut+\frac{1}{2}at^2 }}$

SoLuTiOn :-

Using the first equation,

$\tt{0=10+(-0.5)(t)}\\\\\tt{\implies -10=-0.5t}\\\\\tt{\implies t=\dfrac{-10}{-0.5} }\\\\\boxed{\boxed{\tt{\implies t=20\ s}}}$

__________________

Using the second equation of motion,

$\tt{s=(10)(20)+\dfrac{1}{2}(-0.5)(20)^2 }\\\\\tt{\implies s=200+\dfrac{1}{2}-200 }\\\\\tt{\implies s=200-100 }\\\\\large\boxed{\boxed{\tt{\implies s=100\ m }}}$

∴ The scooter covers a distance of 100 m before it stops.