Solve
Study
Textbooks
Guides
Use app
Login
>>
Class 11
>>
Physics
>>
Motion in a Straight Line
>>
Medium Questions
Motion In A Straight Line
Physics
4727 Views
Easy Qs
Med Qs
Hard Qs
>
Two cars are travelling towards each other on a straight road at velocities
$15m/s$
and
$16m/s$
respectively. When they are
$150m$
apart, both the drivers apply the brakes and the cars decelerate at
$3m/s_{2}$
and
$4m/s_{2}$
until they stop. Separation between the cars when they come to rest is :
A
$86.5m$
B
$89.5m$
C
$85.5m$
D
$80.5m$
Medium
View solution
>
The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is numerically equal to:
A
$g$
B
$2g $
C
$2 g $
D
$2 g$
Medium
View solution
>
A driver takes
$0.20s$
to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of
$54km/h$
and the brakes causes a deceleration of
$6.0m/s_{2}$
, find the distance traveled by the car after he sees the need to put the brakes on.
A
$18.63m$
B
$20m$
C
$26.85m$
D
$27.67m$
Medium
View solution
>
For a body moving with uniform acceleration
$a$
, initial and final velocities in a time interval
$t$
are
$u$
and
$v$
respectively. Then, its average velocity in the time interval
$t$
is :
A
$(v+2at )$
B
$(v−2at )$
C
$(v−at)$
D
$(u+2at )$
Medium
View solution
>
A ball dropped from one metre above the top of a window, crosses the window in
$t_{1}s$
. If the same ball is dropped from
$2m$
above the top of the same window, time taken by it to cross the window is
$t_{2}s$
. Then
A
$t_{2}=t_{1}$
B
$t_{2}=2t_{1}$
C
$t_{2}>t_{1}$
D
$t_{2}<t_{1}$
Medium
View solution
>
A bus starts from rest and moves with a uniform acceleration of
$1ms_{−2}$
. A boy
$10m$
behind the bus at the start runs at a constant speed and catches the bus in
$10s$
. Speed of the boy is :
A
$10ms_{−1}$
B
$1ms_{−1}$
C
$6ms_{−1}$
D
$4ms_{−1}$
Medium
View solution
>
A train accelerates from rest at a constant rate
$a_{1}$
for distance
$S_{1}$
and time
$t_{1}$
. After that it decelerates to rest at a constant rate
$a_{2}$
for distance
$S_{2}$
at time
$t_{2}$
. Then the correct relation among the following is
A
$S_{2}S_{1} =a_{2}a_{1} =t_{2}t_{1} $
B
$S_{2}S_{1} =a_{1}a_{2} =t_{2}t_{1} $
C
$S_{2}S_{1} =a_{2}a_{1} =t_{1}t_{2} $
D
$S_{2}S_{1} =a_{1}a_{2} =t_{1}t_{2} $
Medium
View solution
>
A proton in a uniform electric field moves along a straight line with constant acceleration starting from rest. If it attains a velocity
$4×10_{3}$
$km/s$
at a distance of
$2cm$
, the time required to reach the given velocity is :
A
$10_{−3}s$
B
$10_{−6}s$
C
$10_{−8}s$
D
$10_{−5}s$
Medium
View solution
>
From a building two balls A & B are thrown such that A is thrown upwards and B downwards (both vertically with same speed). If
$V_{A}$
and
$V_{B}$
are their respective velocities on reaching the ground then :
A
$V_{B}<V_{A}$
B
$V_{A}=V_{B}$
C
$V_{A}<V_{B}$
D
Their velocities depends on their masses
Medium
View solution
>
The average velocity of a body moving with uniform acceleration after travelling a distance of
$3.06m$
is
$0.34m/s$
. The change in velocity of the body is
$0.18m/s$
. During this time, its acceleration is
A
$0.01m/s_{2}$
B
$0.02m/s_{2}$
C
$0.03m/s_{2}$
D
$0.04m/s_{2}$
Medium
View solution
>