DC Pandey Solutions (JEE Advance): Motion in One Dimension- 4 | DC Pandey Solutions for JEE Physics PDF Download

Match the Columns

Ques 1: Match the following two columns.

Column I	Column II
	(p) speed must be increasing
	(q) speed must be decreasing
	(r) speed may be increasing
	(s) speed may be decreasing

Ans: (a) → r,s (b) →  r,s (c) →  p (d) →  q
Sol: 
(a)

∴ v must be increasing with time.
⇒ speed must also be increasing with time.
(c) → (p).
(d) Slope is + ive and decreasing.
∴ Velocity must be decreasing with time.
⇒ speed must also be decreasing with time. (d) → (q).

Ques 2: Match the following two columns.

Column I	Column II
	(p) speed increasing
	(q) speed decreasing
	(r) speed constant
	(s) Nothing can be said

Ans: (a) → P (b) → p (c) → q (d) → q
Sol:





 


Speed increasing.
∴ (a) → (p).



∴
∴
As t > 0,  



or
∴
At t = 0 s  

Ques 3: The velocity-time graph of a particle moving along X-axis is shown in figure. Match the entries of Column I with entries of Column II.

Ans: (a) → P (b) → p (c) → q (d) → r
Sol: (a) From A to B, v is increasing and area is + ive.
∴ (a) → (p).
(b) From B to C, v is increasing and area is + ive.
∴ (b) → (p).
(c) From C to D, v is decreasing while area is + ive.
∴ (c) → (q).
(d) From D to E, v is - ive and is
increasing, area is -ive.
∴ (d) → (r).

Ques 4: Corresponding to velocity-time graph in one dimensional motion of a particle is shown in figure, match the following two columns.

Ans: (a) → (r) (b) → (s) (c) → (r) (d) → (r)
Sol: (a) Displacement = Area

i.e., (a) → (r).

= 5.83 unit
∴(b) → (s)


∴ (c) → (s)
(d) Rate of change of velocity at t = 4 s

∴ Rate of change of speed at
t = 4 s would be 5.
i.e., (d) ∴ (r).

Ques 5: A particle is moving along x-axis. Its x-coordinate varies with time as: x = -20 + 5t² 
For the given equation match the following two columns

Ans: (a) → (r) (b) → (q) (c) → (s) (d) → (p)
Sol: (a) x = - 20 + 5t² 
0 = - 20 + 5t²
⇒ t = 2 s
i.e., (a) → (r).
(b) x = - 20 + 5t²

Velocity will be numerically equal to acceleration at t = 1 s.
Thus (b) → (q).
(c) At t = 0 s   x = - 20 m
t = 1 s              x = - 15 m
t = 2 s             x = 0 m
t = 3 s             x = + 25 m
Particle is always moving along + ive x-axis.
∴ (c) → (s).

∴ (d) → (p).

Ques 6: x and y-coordinates of particle movingin x - y plane are,
x = 1 - 2t + t² and y = 4 - 4t + t²
For the given situation match the following two columns.

Ans: (a) → (q) (b) → (p) (c) → (s) (d) → (s)
Sol: (a) x = 1 - 2t + t² 
0 = 1 - 2t + t² [for particle to cross y-axis]
i.e., t = 1 s
Y = 4 - 4t + t²

= - 4 + (2 x 1) [at t = 1 s]
= - 2 unit
∴ (a) → (q).
(b) Y = 4 - 4t + t² 
0 = 4 - 4t + t² [For particle to cross x-axis]
i.e., t = 2 s
x = 1 - 2t + t²

= - 2 + (2 x 2) [at t = 2 s]
= + 2 unit

∴  Initial velocity of particle

i.e., (c) → (s).

and  
∴   Initial acceleration of particle = 2√2 unit
i.e., (d) → (s).