Motion In a Straight Line-Quiz-16

Physics Quiz-16

Linked Comprehension Type

Dear Readers,

If an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. Motion in a straight line is nothing but linear motion. As the name suggests, it’s in a particular straight line, thus it can be said that it uses only one dimension.

Q1. 

Relative velocity of ground w.r.t. B is

•  25 ms^(-1)due north
•  25 ms^(-1)due south
•  40 ms^(-1)due north
•  40 ms^(-1)due south

Solution

There is a balloon containing a stone at a height 300 m from the ground.

(Take g=10ms^(-2))

Q2. 

A stone is dropped from the balloon when balloon is stationary at height 300m. How long will the stone take to reach the ground?

•  7.12s
•  7.75s
•  7.82s
•  8.12s

Solution

t=√(2s/g)=√((2×300)/10)=7.75s

The displacement of a body is given by 4s=M+2Nt^4, where M and N are constants.

Q3. 

The velocity of the body at any instant is

•  (M+2Nt^4)/4
•  2N
•  (M+2N)/4
•  2Nt^3

Solution

s=(M+2Nt^4)/4  ⇒v=ds/dt=2 Nt^3

Putting t=1 s, we get v=2 N

A body is dropped from the top of the tower and falls freely.

Q4. 

The distance covered by it after  second is directly proportional to

• 
   n^2
•  n
•  2n-1
•  2n^2-1

Solution

S=1/2 gn^2  ⇒S∝n^2

S=a/2 (2n-1)⇒S∝(2n-1)

V=gn⇒v∝n

A body at rest is acted upon by a constant force (it means acceleration of the body will be constant).

Q5.

What is the nature of the displacement-time graph?

•  Straight line
•  Parabola
•  Asymmetric parabola
•  Rectangular hyperbola

Solution

Here acceleration is constant. So we can use

s=ut+1/2 at^2,s-t graph will be parabolic

A car accelerates from rest at a constant rate α for some time and then decelerates at a constant rate β to come to rest. The total time elapsed is t

Q6.  

The maximum velocity attained by the car is

•  αβ/(2(α+β)) t
•  αβ/(α+β) t
•  2αβ/(α+β) t
•  4αβ/(α+β) t

Solution

From A to B, applying v=u+at, we get

v_0=0+at_1⇒t_1=v_0/a

From B to C, again applying v=u+at, we get

0=v_0-bt_2⇒t_2=v_0/b

Given t_1+t_2=t ⇒v_0/α+v_0/β=t⇒v_0  αβt/(α+β)

v_0 is the maximum velocity attained

A body is moving with uniform velocity of 8 ms^(-1). When the body just crossed another body, the second one starts and moves with uniform acceleration of 4 ms^(-2)

Q7.

The time after which two bodies meet will be

•  2s
•  4s
•  6s
•  8s

Solution

Let they meet after time t, then the distance travelled by both in time t should be same

s=8t=1/2 4t^2⇒t=4 s

A body is allowed to fall from a height of 100 m. If the time taken for the first 50 m is t_1 and for the remaining 50 m is t_2

Q8. 

Which is correct?

•  t_1=t_2
•  t_1>t_2
•  t_1<t_2
•  Depends upon the mass

Solution

s=1/2 gt_1^2 or  t_1^2=(50×2)/g=100/g  or t_1=10/√g

and 100=1/2 gt^2  or t=(10√2)/√g

t_2=t-t_1=10/√g (√2-1)=0.4t_1

t_1>t_2

A body is dropped from a balloon moving up with a velocity of  4 ms^(-1) when the balloon is at a height of 120.5 m from the ground

Q9.

The height of the body after 5 s from the ground is (g=9.8 m/s^(-2))

• 
   8s
•  12s
•  18s
•  24s

Solution

s=ut+1/2 at^2=4×5-1/2×9.8×5^2

=20-122.5=-102.5 m

This shows that the body is 102.5 m below the initial position, i.e., 

height of the body =120.5-102.5=18 m

A bus starts moving with acceleration 2 ms^(-2). A cyclist 96 m behind the bus starts simultaneously towards the bus at a constant speed of 20 ms^(-1)

Q10. 

After what time will he be able to overtake the bus?

•  4 s
•  8 s
•  12 s
•  16s

Solution

Let at time t, the cyclist overtake the bus, then 96+ (distance travelled by bus in time t)= (distance travelled by cyclist in time t)

⇒1/2×2×t^2+96=20×t⇒t^2-20t+96=0

This gives t=8 s or 12 s. Hence, the cyclist will overtake the bus at 8 s