## Physics Quiz-2

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. From a high tower, at time t=0, one stone is dropped from rest and simultaneously another stone is projected vertically up with an initial velocity. The graph of distance S between the two stones plotted against time t will be
Solution
At time t, let displacement of first stone be S_1=1/2 gt^2 and that of second stone be S_2=ut-1/2 gt^2
Distance between two stones at time t:
S=S1+S2=ut ⇒S=ut

Q2.The displacement x of a particle moving in one dimension under the action of a constant force is related to time t by the equation t=√x+3, where x is in meters and t is in seconds. Find the displacement of the particle when its velocity is zero
•  Zero
•  12 m
•  6 m
•  18 m
Solution
t=√x+3
Differentiating will respect to t, we get
1=1/(2√x)  dx/dt+0 or   dx/dt=2√x
When velocity is zero, 2√x=0 or x=0

Q3.  The x and y coordinates of a particle at any time t are given by: x=7t+4t^2 and y=5t, where x and y are in meters and t in seconds. The acceleration of the particle at 5 s is
•  Zero
•  8 ms^(-2)
•  20 ms^(-2)
•  40 ms^(-2)
Solution
a_x=(d^2 x)/(dt^2 )=8 and a_y=(d^2 y)/(dt^2 )=0
Hence, net acceleration is √(a_x^2+a_y^2 )=8 ms^(-2)

Q4. A particle starts from the origin with a velocity of 10 ms^(-1) and moves with a constant acceleration till the velocity increases to 50 ms^(-1). At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?
• Zero
•  10 ms^(-1)
•  50 ms^(-1)
•  70 ms^(-1)
Solution
2ax=(50)^2-(10)^2 and 2(-a)(-x)=v^2-(50)^2
This given v^2-(50)^2=(10)^2 i.e., v=70 ms^(-1)

Q5.Which graph represents uniform motion?

Solution
Uniform motion involves equal distances covered in equal time intervals or the velocity is constant.

Q6.
 A stone is dropped from rising balloon at a height of 76 m above the ground and reaches the ground in 6 s. What was the velocity of the balloon when the stone was dropped?  Take g=10 ms^(-2)
•  52/3 ms^(-1) upward
•  (52/3) ms^(-1) downward
• 3 ms^(-1)
•  9.8 ms^(-1)
Solution
S=ut+1/2 at^2
⇒ -76=u×6-1/2×10×(6)^2⇒u=52/3  ms^(-1)

Q7.
The distance travelled by a particle in a straight line motion is directly proportional to t^(1/2), where t=        time elapsed. What is the nature of motion?
•  Increasing acceleration
•  Decreasing acceleration
•  Increasing retardation
•  Decreasing retardation
Solution

s=kt^(1/2)  ⇒a=(d^2 s)/(dt^2 )=-1/4 kt^(-3/2)
As t increases, retardation decreases

Q8. From the velocity-time graph, given in Fig of a particle moving in a straight line, one can conclude that

•  Its average velocity during the 12 s interval is 24/7 ms^(-1)
•  Its velocity for the first 3 s is uniform and is equal to 4 ms^(-1)
• The body has a constant acceleration between t=3 s and t=8 s
•  The body has a uniform retardation from t=8 s to t=12 s
Solution
Displacement in 12 s = area under v-t graph
=1/2×(12+5)4=34 m
Vav=Displacement/Time=34/12=17/6  ms^(-1)
Hence (a) is incorrect; (b) is incorrect because during first 3 seconds, velocity increases from 0 to 4 ms^(-1) option; (c) is incorrect, because in part AB velocity is constant

Q9.
 Plot the acceleration-time graph of the velocity-time graph given in Fig Solution
For 0 to 5 s, acceleration is positive, for 5 to 15 s acceleration is negative, for 15 to 20 s acceleration is positive.

Q10. A stone is dropped from a certain height which can reach the ground in 5 s. It is stopped after 3 s of its fall and then it is again released. The total time taken by the stone to reach the ground will be
•  6s
•  6.5s
•  7s
•  7.5s
Solution
Here h=1/2×10×(5)^2=125 m
In 3 s it falls through: h_1=1/2×10×(3)^2=45 m
Rest 80 m is covered in 4 s. Hence, total time taken is 3 s+4 s=7 s #### Written by: AUTHORNAME

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