A train moves from one station to another in two hours time it's speed during the motion. determine the maximum acceleration during the whole journey. also calculate the distance covered during the time interval from 0.75 to 1 hr.

Open in App

Solution

(i) Since acceleration is the rate of change of velocity, it is given by the slope of the velocity time curve. Here the slope is greatest between ¾ and 1 hour.

Change in the velocity in this interval = (60-20) km h⁻¹

.·. Acceleration in this interval = (60-20)/(1-(3/4)) km h⁻²

= 160 km h⁻²

(ii) Distance travelled during this interval is given by $s = u t + 1 / 2 a t ²$

$= 20 (1 / 4) + 1 / 2 (160) (1 / 4) ²$

$= 10 k m$

Suggest Corrections

Similar questions

Q. A train moves from one station to another in

2

hours time, its speed during the motion is shown in graph above. The maximum acceleration during the journey and distance covered during the time interval from

0.75

hour to

1.00

hour are

Q. The total distance between two stations is

570

km out of which ,

224

km is covered at

56

km/hr,

210

km is covered at

70

km/hr and remaining is covered in

2

hours. Find the average speed during the whole journey.

Distance between two stations Vizag and Bangalore is 780 km. A train covers the journey from Vizag to Bangalore at 86 km per hour and returns back to Vizag with a uniform speed of 65km per hour. Find the average speed of the train during the whole journey?

A body starts moving with an initial velocity 50 m/s and acceleration 20 $m / s^{2}$ . How much distance it will cover in 4s ? Also, calculate its average speed during this time interval.

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.

Join BYJU'S Learning Program