The acceleration of a train between two stations is shown in the -gure- The maximum speed of the train is

The correct option is D 30 m/sv_max = Maximum positive area of a t graph = area between 0 and 8s = 30 m/s ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Acceleration

The accelerat...

Question

The acceleration of a train between two stations is shown in the ﬁgure. The maximum speed of the train is

60 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 m/s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

120 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

90 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 30 m/s
$v_{m a x}$ = Maximum positive area of a-t graph
= area between 0 and 8s
= 30 m/s

Suggest Corrections

Similar questions

If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/ $s^{2}$ and subway stations are located 880 m apart. What is the maximum speed, a subway train can attain between stations?

Q. A train leaves station 'A' to station 'B'. The train travels straight without any halts between the stations. During the first and last

200

m of its journey, the train has uniform acceleration and retardation both equal to

1 {m s}^{- 1}

respectively. For the rest of the journey, the train maintains uniform speed Calculate the average speed of the train, given the distance between two stations is 4 km.

A train and a car start together from a point. The train's speed is 80 kmph and the car's speed is 60% of the train's speed. The train crosses a station in 15 minutes. How much distance will the car travel in the time the train takes to cross 3 stations? (Assume constant distance between consecutive stations)

Q. A train starts from rest from a stations with acceleration

0.2 m / s^{2}

on a straight track and then comes to rest after attaining maximum speed on another station due to retardation

0.4 m / s^{2}

.IF total time spent is half an hour,then distance between two stations is [Neglect length of train]

Between two stations a train accelerates uniformly at first then moves with constant velocity and then retards uniformly. If the ratio of the time taken is1:8:1,and the maximum speed attained by the train is 60 km per hour ,then what is the average speed of the whole journey?

Join BYJU'S Learning Program

The acceleration of a train between two stations is shown in the ﬁgure. The maximum speed of the train is

The correct option is B 30 m/svmax = Maximum positive area of a-t graph = area between 0 and 8s = 30 m/s

The correct option is B 30 m/s
$v_{m a x}$ = Maximum positive area of a-t graph
= area between 0 and 8s
= 30 m/s