A particle starts moving from rest state along a straight line under the action of a constant force and travel distance x in first 5 sec. The distance travelled by it in next five seconds will be
Answers
From second law of motion:
S=ut+1/2at²
u=0 m/s
a= am/s²
time=5sec
Let x be the distance travelled by particle.
X=0*t+1/2xax5x5
=25a/2
a=2x/25
Final velocity :
V=u+at
v=at=(2x/25)xt
=2x/5
Distance travelled in next five seconds:
s1=ut+1/2at²
s1=(2x/5)x5+1/2(2x/25)x5²
=2x+x
=3x
∴The distance travelled by it in next five seconds will be 3x m
GIVEN :
A particle starts moving from rest state along a straight line under the action of a constant force and travel distance x in first 5 sec.
TO FIND :
The distance travelled by it in next five seconds will be
SOLUTION :
◆From rest , due to constant force particle travelled x meters.
◆Constant force denotes constant acceleration, as mass is constant ( this is not a relative motion)
◆Einstein's second law ,
S = u t + 1/2at²
(u- initial velocity , a - acceleration. t- time)
F = ma .
U = 0 m/s . Since it's started from rest.
t=5sec , acceleration, a ms^-2 ,
x - distance travelled (given)
◆Substituting these values on equation,
◆x =0×t + 1/2 x a x 5 x 5
=12.5a
◆a=2x/25
◆Einstein's first law
Final velocity ,v =u+at
◆v = at = (2x/25) x t (as u=0)
= (2x/25 ) × 5
= 2x/5.
◆Distance travelled in next 5 second is,
s' = ut + 1/2at²
◆Initial velocity u is final velocity ,v travelled in first 5 seconds
◆On substitution,
◆s' = (2x/5) x 5 + 1/2 (2x/25) x 5²
= 3x.
ANSWER :
3x meters distance is travelled by particle in next five seconds .