# Motion in straight Line

## Motion in straight Line Solved Examples Part 3

13. A body fall from height H.if t1 is time taken for covering first half height and
t2 be time taken for second half.Which of these relation is true for t1 and t2
a. t1 > t2
b. t1 < t2
c. t1=t2
d. Depends on the mass of the body

Solution(13):
. Let H be the height
then
First Half
H/2=(1/2)gt12 ----(1)
or
(1/2)gt12=H/2
Also v=gt1

Second Half
H/2=vt2+(1/2)gt22
or
(H/2)=gt1t2+(1/2)gt22
or
(1/2)gt22 =(H/2)-gt1t2 ---(2)

From 1 and 2
(1/2)gt22=(1/2)gt12 -gt1t2
t22+2t1t2 -t12=0

or t2=[-2t1+√(4t1t2 +4t1t2 )]/2
or t2=[-2t1+2t1√8]/2
t2=.44t1

so t1 > t2

Hence a is correct

14.A train running at 30m/s is slowed uniformly to a stop in 44 sec.Find the stopping distance?
a. 612 m
b. 662 m
c. 630 m
d. 605 m

Solution(14):
. Here
v0=30m/s vf=0 at t=44 sec
vf=v0+at or 0=30+a(44) or a=-.68m/s2

Now x=v0t+(1/2)at2
or x=662 m

Hence b is correct

15.A nut comes loose from a bolt on the bottom of an elevator as the elavator is moving up the shaft at 3m/s.The nut strikes the bottom of the shaft in 2 sec.How far from the bottom of the shaft was the elevator when nut falls off?
a. 13.6 m
b. 10 m
c. 12.6 m
d. none of these

Solution(15):
. Here the nut intially has the velocity of the elevator ,so choosing upward as positive ,v0=3 m/s
also a=-g=-9.8 m/s2

The time taken to hit the bottom is 2 s so
y=v0t+(1/2)at2
=-13.6 m

Hence the bottom of the shaft was 13.6 m below elevator when nut fall off
Hence (a) is correct
16. Let v and a denote the velocity and acceleration respectivly of the particle in one dimension motion.
a.the speed of the particle decreases when v.a <0
b. the speed of the particle decreases when v.a >0
c.the speed of the particle increases when v.a=0
d. the speed of the particle decreases when |v|<|a|

Solution(16):
. when v.a <0
That mean they are in opposite direction
So when the particle is moving towards origin,acceleration is acting outwards so it is decreasing the speed
When the particle is moving outwards,acceleration is acting inwards hence it is decreasing the speed
So this option is correct

When v.a > 0
That means v and a are in same direction
When the particle is moving towards origin, acceleration is also acting inwards so increasing the speed
When the particle is moving outwards,acceleration is also acting outwards hence increasing the speed
So this option is not correct

when v.a=0
Then possible cases are
Acceleration is zero
Velocity is zero
Both are zero
So speed does not increases in all cases.So this option is not correct

When
|v|<|a|
Possible cases
Velocity is positive ,accleration is positive ----Speed increase
Velocity is negative ,acceleration is negative----Speed increase
Velocity is positive,acceleration is negative
Velocity is negative ,acceleration is positive---Speed decreases
So this is not correct option