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Question: A bus starting from rest moves with a uniform acceleration of for 2 minutes, find (a) the speed acqu...

A bus starting from rest moves with a uniform acceleration of for 2 minutes, find (a) the speed acquired (b) the distance travelled.

Explanation

Solution

As in the given problem,the acceleration is uniform or we can say it is constant so we use Newton’s equation of motion for calculating final speed and distance.

Formula used: Newton’s I, equation is
v=u+at\vec v = \vec u + \vec at
Where v=\vec v = Final velocity, u=\vec u = Initial velocity, t=t = time
Newton’s II equation is
S=u+12at2\vec S = \vec u + \dfrac{1}{2}\vec a{t^2}
Where S == displacement travelled by particle
Newton’s III equation is
v2+u2+2as{v^2} + {u^2} + 2\vec a \cdot \vec s

Complete step by step answer:
Given that (acceleration)a=0.1m/s2(acceleration)a = 0.1m/{s^2}
Because the initial bus is at rest. So, initial velocity u=0u = 0
Time t == 2 minutes
t=2×60sect = 2 \times 60\sec
t=120sect = 120\sec
i. We have to find the final speed. So, applying Newton’s I equation
v=u+atv = u + at
v=0+(0.1)(120)v = 0 + (0.1)(120)
v=12m/sv = 12m/s
Hence final speed of bus after 2 min is 12m/s12m/s
ii. Now for calculating distance travelled we use Newton’s II equation
S=ut+12at2S = ut + \dfrac{1}{2}a{t^2}
S=(0)×(120)+12×(0.1)×(120)2S = (0) \times (120) + \dfrac{1}{2} \times (0.1) \times {(120)^2}
S=0+12×0.1×120×120S = 0 + \dfrac{1}{2} \times 0.1 \times 120 \times 120
S=12×1202S = \dfrac{{12 \times 120}}{2}
S=6×120S = 6 \times 120
S=720mS = 720m
Hence, the distance travelled by the bus in 2 min is 720 m.

Note: In order to solve these type of problems, we always should remember that where and which equation of Newton’s we should apply for e.g.if acceleration time and initial velocity is given and we have to find final velocity of body then we should apply Newton’s I equation. If displacement is asked in Numerical then we should apply Newton’s II equation.