Ask a Teacher

Search



A CAR STARTS FROM REST AND ACCELERATE UNIFORMLY FOR 10 SECONDS TO A VELOCITY OF 8M/S.IT THEN RUNS WITH CONSTANT VELOCITY AND IS FINALLY GO TO REST IN 64m WITH CONSTANT RETARDATION. THE TOTAL DISTANCE COVERED BY THE CAR IS 584m.FIND THE VALUE OF ACCELERATION,RETARDATION AND TOTAL TIME TAKEN.
1. Using the equations of the accelerated motion we have:

v = a*t  
s = (a*t^2)/2 
Given t = 10s , v = 8m/s ,u=0
a = v/t 
  = 0.8 m/s^2
s = 40 m.

2.  During retardation

s = 64 m  
v = 8 m/s
using formula: r = (v^2)/(2*s)   
retardation r = 0.5 m/s^2
v = t*r 
t = v/r
  =16 s

3.  With uniform velocity
v = 8 m/s  
 a=0
s = 584 - 40 - 64 
  = 480m
t = s/v 
  = 60 s

So, total time T = 10+16+60 = 86 s


comments powered by Disqus