Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
If the greatest admissible acceleration of a train is 3 feet/sec square, calculate the least time taken from one station to another at a distance of 10m.
Maximum permissible acceleration = 0.9144 m/s.
If initial velocity=0;
t=((2x10)/0.9144)1/2
=4.67 sec.
i assume that the train starts from the first station and crosses the second station with the same acceleration.
so the initial velocity is zero.
the acceleration is 0.9m/s2.
time taken is (2*10/0.9)(1/2) = 4.71 s.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !