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
|----------------|---------------|
|<------x----->|<---200-x--->|
let us assume distance travelled during accn is x and time taken is t1 nd distance travelled during retardation is 200-x and time taken is t2......then for accn we can write newtons eqution as---
v=u+at1
u=0 initially at rest
thus v=t1......a=1m/s2------(1)
now applying newtons second law of motion .....
s=ut+1/2(at2)
x=(1/2)t12....as u=0,a=1 .....
v=t1=√2x........(2)
now considering the retardation path and newtons (3) law of motion
v2-u2=2as v=u+at.......v=0.....u=√2x....a=-4........t2=(√2x)/4.....(3)
as we know initial velocity will be the final velocity of acceleration path and final velocity will be zero thus
-u2=-8(200-x)....as a=-4m/s2
(√2x)2=2(800-4x)....as u=√2x
x=800-4x
5x=800.................x=160..........t1=√2*160(from eqn 2)
thus t1=8√5
t2=2√5....from eqn 3
total time=t1+t2=10√5=22.5sec......answer
if ur sattisfied with the reply then click on 'yes'...
similarly, you can chech other case that v1 reaches 24m/s dur. acc. but time reqd. will come to be more.
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 !