A train accelerating uniformly from rest attains a maximum speed of 40 ms^ -1 in 20 s . It travels at the speed for 20 s and is brought to rest with uniform retardation in further 40 s . What is the average velocity

Correct answer: B. 25 m / s (i) v=u+ at _ 1 array l 40=0+a × 20 a=2 m / s^ 2 v^ 2 -u^ 2 =2 a s 40^ 2 -0=2 × 2 s_ 1 (ii) s_ 1 =400 m (iii) v=u+a t 0=40+a × 40 a=-1 m / ^ 2 0^ 2 -40^ 2 =2(-1) s_ 3 s_ 3 =800 m Total distance travelled =s_ 1 +s_ 2 +s_ 3 =400+800+800=2000 m Total time taken =20+20+40=80 s Average velocity =(2000)/(80)=25 m / s array

BITSAT 2014

Exam: BITSAT
Year: 2014
Subject: Physics
Chapter: Motion in One Dimension

BITSAT 2014 · Physics · Motion in One Dimension

A train accelerating uniformly from rest attains a maximum speed of...

A train accelerating uniformly from rest attains a maximum speed of

40 \mathrm{ ms}^{-1}

20 \mathrm{ s}

. It travels at the speed for

20 \mathrm{ s}

and is brought to rest with uniform retardation in further

40 \mathrm{ s}

. What is the average velocity during the period ?

Options

A $80 \mathrm{ m} / \mathrm{s}$
B $25 \mathrm{ m} / \mathrm{s}$
C $40 \mathrm{ m} / \mathrm{s}$
D $30 \mathrm{ m} / \mathrm{s}$

Correct answer

B. $25 \mathrm{ m} / \mathrm{s}$

Step-by-step solution

(i)

\mathrm{v}=\mathrm{u}+

at _{1}

\begin{array}{l} 40=0+a \times 20 \\ a=2 m / s^{2} \\ v^{2}-u^{2}=2 a s \\ 40^{2}-0=2 \times 2 s_{1} \\ \text{ (ii) } s_{1}=400 \mathrm{ m} \\ \text{ (iii) } v=u+a t \\ \quad 0=40+a \times 40 \\ a=-1 \mathrm{ m} / \mathrms^{2} \\ 0^{2}-40^{2}=2(-1) \mathrm{s}_{3} \\ \mathrm{ s}_{3}=800 \mathrm{ m} \\ \text{ Total distance travelled }=\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{3} \\ =400+800+800=2000 \mathrm{ m} \\ \text{ Total time taken }=20+20+40=80 \mathrm{ s} \\ \text{ Average velocity }=\frac{2000}{80}=25 \mathrm{ m} / \mathrm{s} \end{array}

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