Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1: 8: 1 and the maximum speed attained be 60 kmh -1, then what is the average speed over the whole journey?

Question

Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1: 8: 1 and the maximum speed attained be 60 kmh -1, then what is the average speed over the whole journey? (A) 48 kmh -1 (B) 52 kmh -1 (C) 54 kmh -1 (D) 56 kmh -1

Tardigrade Admin · Accepted Answer

S1=(1/2) a t2=(1/2)(a t) t=(60 t/2)=30 t S2=60 × 8 t=480 t, S3=S1=30 t <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/73f0eb21b7b47dafddd123c5cae61751-.png" /> v av =(S1+S2+S3/t+8 t+t) =54 km h -1

$48 km h^{- 1}$

$52 km h^{- 1}$

$54 km h^{- 1}$

$56 km h^{- 1}$

$S_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} (a t) t = \frac{60 t}{2} = 30 t$
$S_{2} = 60 \times 8 t = 480 t, S_{3} = S_{1} = 30 t$

$v_{a v} = \frac{S _{1} + S _{2} + S _{3}}{t + 8 t + t}$
$= 54 km h^{- 1}$

Q. Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1:8:1 and the maximum speed attained be 60kmh−1, then what is the average speed over the whole journey?

48kmh−1

52kmh−1

54kmh−1

56kmh−1

S1​=21​at2=21​(at)t=260t​=30t S2​=60×8t=480t,S3​=S1​=30t vav​=t+8t+tS1​+S2​+S3​​ =54kmh−1

$48 km h^{- 1}$

$52 km h^{- 1}$

$54 km h^{- 1}$

$56 km h^{- 1}$

$S_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} (a t) t = \frac{60 t}{2} = 30 t$
$S_{2} = 60 \times 8 t = 480 t, S_{3} = S_{1} = 30 t$

$v_{a v} = \frac{S _{1} + S _{2} + S _{3}}{t + 8 t + t}$
$= 54 km h^{- 1}$