Q. 24.2( 98 Votes )

A train is traveling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of, -0.5 ms-2. Find how far the train will go before it is brought to rest.

Answer :

Initial speed, u = 90 km/h2

(Converting it into m/s, We know 1km=1000m and 1 hour= 3600 seconds)


= 25 m/s

Final speed, v = 0 (As the train stops)

Acceleration, a = -0.5 m/s2(As the brakes are being applied the speed is reducing)


Distance traveled, s =? (To be Calculated)

Now, v2 = u2 + 2as

(0)2 = (25)2 + 2 × (-0.5) × s

0 = 625 – 1 × s
0 = 625 - s

∴s = 625 m (Taking s on the other side will make it positive)

Therefore, 625 meters is the distance traveled by the train before it is brought to rest.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Champ Quiz | Motion under gravity46 mins
About Acceleration50 mins
Velocity Time Graph46 mins
Second Equation of Motion48 mins
Understanding Motion60 mins
Interactive Quiz - Motion29 mins
Learn The Use of Equations of Motion37 mins
Learning Second Equation of motion41 mins
Third Equation of Motion36 mins
NCERT | Motion - Part 153 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses