Question

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

Answer 1

Answer:

u= 43.2 km/h = 43.2 ×5/3 = 72 m/s

a= -6

v= 0

${v}^{2} = {u}^{2} + 2as \\ \\ \\ 0 = {72}^{2} - 2 \times 6 \times s \\ \\ 5184 = 12s \\ \\ s = \frac{5184}{12 } = 432$

Answer 2

Answer:

12 m

Explanation:

We have,

• initial velocity (u) = 43.2 km/h = 12 m/s
• acceleration (a) = -6 m/s²
• final velocity (v) = 0 m/s

We know that,

$\underline{\boxed{\sf v^2=u^2+2as}}$

$\sf \longmapsto 0 = (12)^2+2(-6)(s)$

$\sf \longmapsto 0=144-12s$

$\sf \longmapsto -12s=-144$

$\sf \longmapsto s = \dfrac{144}{12}=12 \: m$