# Can anyone confirm the answer I have to this question?

## Total time: 225 sec Total distance travelled: 8100 m Speed: 36 m/s

Feb 4, 2017

Given

For first phase journey

${u}_{1} \to \text{initial velocity of the car } = 0$
${t}_{1} \to \text{duration of first phase journey} = 15 s$
${a}_{1} \to \text{uniform acceleration in first phase} = 2.4 m {s}^{-} 2$

So applying kinematic equation $s = u t + \frac{1}{2} a {t}^{2}$
we get distance traversed in first phase

${s}_{1} = {u}_{1} \times {t}_{1} + \frac{1}{2} \times {a}_{1} \times {t}_{1}^{2}$

$\implies {s}_{1} = 0 \times 2.4 + \frac{1}{2} \times 2.4 \times {15}^{2} = 270 m$

For Second phase journey

At the end first phase it will acquire velocity ${v}_{2}$ which will be initial velocity for second phase journey.

So ${v}_{2} = {u}_{1} + {a}_{1} \times {t}_{1} = 0 \times {t}_{1} + 2.4 \times 15 = 36 m {s}^{-} 1$

${t}_{2} \to \text{duration of second phase journey} = 200 s$

This journey occurs at constant velocity ${v}_{2}$

So distance traversed in this phase

${s}_{2} = {v}_{2} \times {t}_{2} = 36 \times 200 = 7200 m$

For third phase journey

${v}_{2} \to \text{initial velocity} = 36 m {s}^{-} 1$

${a}_{2} \to \text{uniform acceleration in 3rd phase} = - 3.6 m {s}^{-} 2$

${v}_{3} \to \text{Final velocity of the car } = 0$

t_3->"duration of 3rd phase journey"=?

So ${v}_{3} = {v}_{2} + {a}_{2} \times {t}_{3}$

$\implies 0 = 36 + \left(- 3.6\right) \times {t}_{3}$

$\implies {t}_{3} = 10 s$

Distance traversed in 3rd phase

${s}_{3} = \frac{1}{2} \left({v}_{2} + {v}_{3}\right) \times {t}_{3}$

$\implies {s}_{3} = \frac{1}{2} \left(36 + 0\right) \times 10 = 180 m$

So total duration $\text{ } T = {t}_{1} + {t}_{2} + {t}_{3} = 15 + 200 + 10 = 225 s$

Total distance traversed

$S = \left({s}_{1} + {s}_{2} + {s}_{3}\right) = \left(270 + 7200 + 180\right) m = 7650 m$

So average speed ${V}_{\text{av}} = \frac{S}{T} = \frac{7650}{225} m {s}^{-} 1 = 34 m {s}^{-} 1$

Feb 4, 2017

$\text{A different approach}$

#### Explanation:

$\text{------------------------------------------------}$

$\text{use triangle OAE(for phase 1)}$

$\tan \alpha = \text{acceleration}$

$2.40 = \frac{v}{15}$

$v = 36 \text{ m/s}$

$\text{area=distance covered}$

$\text{area="(36*15)/2=270" meters}$

$\text{-----------------------------------------------}$

$\text{use rectangle ABFE(for phase 2)}$

$\text{area=distance covered for interval t=15 and t=215}$

$\text{area="200*36=7200" meters}$

$\text{-----------------------------------------------}$

$\text{use triangle BFC(for phase 3)}$

$\tan \beta = \text{deceleration}$

$3.6 = \frac{36}{k}$

$k = \frac{36}{3.6} = 10$

$t = 10 + 215 = 225 \text{ s}$

$\text{area=distance covered for interval t=215 and t=225}$

$a r e a = \frac{36 \cdot 10}{2} = \frac{360}{2} = 180 \text{ meters}$

$\text{time elapsed="15+200+10=225 " seconds}$

$\text{total distance covered="270+7200+180=7650" meters}$

"average speed=" ("total distance covered")/("elapsed time")

$\text{average speed="7650/225=34" m/s}$