A person travelling at 43-2 km - h applies the brake giving a deceleration of 6-0 m - s 2 to his scooter- How far will it travel before stopping-

Initial velocity, u = 43.2 km/h = 12 m/s Final velocity, v = 0 Acceleration, a = minus;6 m/s2 From v2=u2+2as, we get: Distance, s=v2 u22(a) #8658;s=0 1222 ...

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Standard IX

Physics

Derivation of Position-Velocity Relation by Graphical Method

A person trav...

Question

A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s² to his scooter. How far will it travel before stopping?

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Solution

Initial velocity, u = 43.2 km/h = 12 m/s
Final velocity, v = 0
Acceleration, a = −6 m/s²
From $v^{2} = u^{2} + 2 a s$ , we get:
Distance, $s = \frac{v^{2} - u^{2}}{2 (a)}$
$\Rightarrow s = \frac{0 - {(12)}^{2}}{2 (- 6)} = \frac{{(12)}^{2}}{12} = 12 m$

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A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s2 to his scooter. How far will it travel before stopping?

Initial velocity, u = 43.2 km/h = 12 m/s Final velocity, v = 0 Acceleration, a = −6 m/s2 From v2=u2+2as, we get: Distance, s=v2-u22(a) ⇒s=0-1222-6=12212=12 m

A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s² to his scooter. How far will it travel before stopping?

Initial velocity, u = 43.2 km/h = 12 m/s
Final velocity, v = 0
Acceleration, a = −6 m/s²
From $v^{2} = u^{2} + 2 a s$ , we get:
Distance, $s = \frac{v^{2} - u^{2}}{2 (a)}$
$\Rightarrow s = \frac{0 - {(12)}^{2}}{2 (- 6)} = \frac{{(12)}^{2}}{12} = 12 m$