A car starts from rest and moves along the x-axis with constant acceleration 5m/s^2 for 8 second. If it then continues with constant velocity, what distance will the car cover in 12 sec since it started from rest?

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would u mind commenting the basic eqns..??
nops... but the same prob is with me also... not having basic eqns...:(
thanks for marking it as the bst..:)

Answers

2014-06-16T19:34:32+05:30
V=u+at
v=0+5m/s^2*8
v=40m/s
constant velocity=40m/s
distance covered in the first 8 sec = S=1/2a*t^2=160m.
distance covered in the last 4 sec=const. speed*time=40m/s*4=160m
total distance covered in 12 sec=160+160=320m ans.
ans=320m

plzz mark it as the best...

11 4 11
thanx for your answer
little confusion to understand the distance of first 8 seconds
can you make this clear to me ??
The Brainliest Answer!
2014-06-16T22:03:47+05:30
Lets divide the qstn into 2 parts..
(1) the distance it travelled in the frst 8 sec
(2) the distance it travld inte next 4 secs..that is in total 12 sec.
(1) given,
  u=0
a=5m/s^2
t=8s
using the eqn s=ut+1/2 at2,
s=0+(1/2)*5*(8^2)
s=160m..................this s d distnce travelled in 8 sec.

now while considering the 2nd part the final velocity in the frst becomes the initial in the sec.
find the final velocity. v=u+at..(not sure abt dis eqn....sry.)
u get v=40m/s.
in the secnd prt it is said that the body moves with uniform velocity. ie a=0
using the eqn s=ut+1/2at2,
u get s=(40*4)+0
s=160m.
thrfore the total distance travelled equals 160+160
=320m.
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