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 during the period?

Question

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 during the period? (A) 80 m/s (B) 25 m/s (C) 40 m/s (D) 30 m/s

Tardigrade Admin · Accepted Answer

S 1=(( u + v /2)) t =20 × 20=400 m S 2=40 × 20=800 m S 3=(( u + v /2)) t =20 × 40=800 m Total displacement =2000 m Total time taken =80 m V =(2000/80)=25 m / s Total time taken =80 m V =(2000/80)=25 m / s

Q. A train accelerating uniformly from rest attains a maximum speed of $40 m s^{- 1}$ 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 during the period?

80 m/s

25 m/s

40 m/s

30 m/s

$S_{1} = (\frac{u + v}{2}) t$
$= 20 \times 20 = 400 m$
$S_{2} = 40 \times 20 = 800 m$
$S_{3} = (\frac{u + v}{2}) t$
$= 20 \times 40 = 800 m$
Total displacement $= 2000 m$
Total time taken $= 80 m$
$V = \frac{2000}{80} = 25 m / s$
Total time taken $= 80 m$
$V = \frac{2000}{80} = 25 m / s$

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