Physics Quiz-16
Linked Comprehension Type
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.
A car is moving towards south with a speed of 20 ms^(-1). A motorcyclist is moving towards east with a speed of 15 ms^(-1). At a certain instant, the motorcyclist is due south of the car and is at a distance of 50 m from the car
Q1.
The shortest distance between the motorcyclist and the car is
Solution
CM=50 m
v_1=20 ms^(-1),v_2=15 ms^(-1),AC=v_1 t,MB=v_2 t
S=√(MA^2+MB^2 )=√((50-v_1 t)^2+(v_2 t)^2 )
From ds/dt=0 (for minima), find t and put t in s
Consider a particle moving along the x-axis as shown in Fig. Its distance from the origin O is described by the coordinate x which varies with time. At a time t_1, the particle is at point P, where its coordinate is x_1, and at time t_2 it is at point Q, where its coordinate x_2. The displacement during the time interval from t_1 and t_2 is the vector from P to Q the x-component of this vector is (x_2-x_1) and all other component are zero
It is convenient to represent the quantity x_2-x_1, the change in x, by means of a notation using the Greek letter Δ (capital delta) to designate a change in any quantity. Thus we write Δx=x_2-x_1 in which Δx is not a product but is to be interpreted as a single symbol representing the change in the quantity x. Similarly, we denote the time interval from t_1 to t_2 as t=t_2-t_1
The average velocity of the particle is defined as the ratio of the displacement Δx to the time interval Δt. We represent average velocity by the letter v with a bar (v ̅) to signify average value.
Thus
v ̅=(x_2-x_1)/(t_2-t_1 )=Δx/Δt
Q2.
A particle moves half the time of its journey with velocity u. The rest of the half time it moves with two velocities v_1 and v_2 such that half the distance it covers with v_1 and the other half with v_2. Find the net average velocity assume straight line motion
Solution
S_1=ut/2,S_2/2=v_1 t_1=v_2 t_2 (i)
t/2=t_1+t_2
From (i) and (ii) ⇒t_1=(v_2 t)/(2(v_1+v_2))
v_av=S/t=(S_1+S_2)/t=(ut/2+2v_1 t_1)/t
Put the value of t_1 and get v_av=(u(v_1+v_2 )+2v_1 v_2)/(2(v_1+v_2))
Two particles A and B are initially 40 m apart, A is behind B. Particle A is moving with uniform velocity of 10 ms^(-1) towards B. Particle B starts moving away from A with constant acceleration of 2 ms^(-2)
Q3.
The time at which there is a minimum distance between the two is
Solution
Distance between the particles will be minimum when velocity of B becomes equal to that of A, i.e., 10 ms^(-1)
Apply v=u+at⇒10=0+2t ⇒t=5 s
The velocity-time graph of a particle in straight line motion is shown in Fig. The particle starts its motion from origin
Q4.
|
The distance travelled by the particle in 8 s is |
Solution
Distance covered = area of speed – time graph
=1/2×(4+2)×4+1/2 (4+2)×2=18 m
Q5.
|
The value of t |
Solution
Average speed =20 ms^(-1)
(Total distance)/(Total time)=20 ⇒ (Area of graph)/(20+t)=20
⇒1/2 (20+t+20-t)×5t=20(20+t)⇒t=5 s
Study the four graphs given below. Answer the following questions on the basis of these graphs
Q6.
In which of the graphs, the particle has more magnitude of velocity at t_1 than at t_2
Solution
In graph (i) and (iii), magnitude of slope is greater at t_1 than that at t_2
Study the following graphs
Q7.
The particle is moving with constant speed
Solution
For the graph (i) and (iv), slope is constant, hence the velocity is constant
