Kinematics-1D
Q1
Objective
Match the columns
Numerical
Match the following columns and mark the correct option from the codes given below.
| Column I | Column II | ||||||||
| (A) | M | (p) | ${A}^{-1}$ | ||||||
| (B) | N | (q) | ${R}^{-1}$ | ||||||
| (C) | P | (r) | A | ||||||
| (D) | Q | (s) | R | ||||||
| Codes | |||||||||
| A | B | C | D | A | B | C | D | ||
| (a) | p | r | q | s | (b) | r | s | p | q |
| (c) | s | p | r | q | (d) | q | p | s | r |
Correct Answer: B
Explanation:
With constant positive acceleration, speed will increase when velocity is positive, speed will decrease, if velocity is negative.
Similarly, with constant negative acceleration, speed will increase, if velocity is negative and speed will decrease, if velocity is positive.
Hence, $\dot{A} \to \mathrm{p}q$, $\dot{\mathbf{B}} \to \mathrm{p}q$, $\dot{\mathbf{C}} \to \mathbf{r}$, $\mathrm{D} \to \mathrm{p}q$.
Similarly, with constant negative acceleration, speed will increase, if velocity is negative and speed will decrease, if velocity is positive.
Hence, $\dot{A} \to \mathrm{p}q$, $\dot{\mathbf{B}} \to \mathrm{p}q$, $\dot{\mathbf{C}} \to \mathbf{r}$, $\mathrm{D} \to \mathrm{p}q$.
Q2
Objective
Match the columns
Numerical
In the $s - t$ equation $(s = 10 + 20t - 5t^2)$, match the following columns and mark the correct option from the codes given below.
| Column I | Column II | ||||||
| (A) | Distance travelled in 3 s | (p) | -20 units | ||||
| (B) | Initial acceleration | (q) | 15 units | ||||
| (C) | Velocity at 4 s | (r) | 25 units | ||||
| (s) | -10 units | ||||||
| Codes | |||||||
| A | B | C | A | B | C | ||
| (a) | p | q | s | (b) | r | q | s |
| (c) | r | s | p | (d) | q | s | r |
Correct Answer: C
Explanation:
At $ t = 3 \, \mathrm{s} $, $ s = 10 + 20 \, (3) - 5(3)^0 = 25 \, \mathrm{unit} $
$v = \frac {d s}{d t} = 2 0 - 1 0 t$
At $ t = 4 \, \mathrm{s} $, $ v = 20 - 10(4) = -20 \, \mathrm{unit} $
$a = \frac {d v}{d t} = - 1 0 \mathrm{units}$
Hence, $\dot{\mathbf{A}}\rightarrow \mathbf{r},\mathbf{B}\rightarrow \mathbf{s},\mathbf{C}\rightarrow \mathbf{p}.$
$v = \frac {d s}{d t} = 2 0 - 1 0 t$
At $ t = 4 \, \mathrm{s} $, $ v = 20 - 10(4) = -20 \, \mathrm{unit} $
$a = \frac {d v}{d t} = - 1 0 \mathrm{units}$
Hence, $\dot{\mathbf{A}}\rightarrow \mathbf{r},\mathbf{B}\rightarrow \mathbf{s},\mathbf{C}\rightarrow \mathbf{p}.$
Q3
Objective
Match the columns
Numerical
For the velocity-time graph as shown in the figure, in a time interval from t = 0 to t = 6 s, match the following columns and mark the correct option from the codes given below.
| Column I | Column II | ||||
| (A) | Change in velocity | (p) | -5/3 SI unit | ||
| (B) | Average acceleration | (q) | -20 SI unit | ||
| (C) | Total displacement | (r) | -10 SI unit | ||
| (D) | Acceleration at $t = 3$ s | (s) | -5 SI unit | ||
| Codes | |||||
| A | B | C | D | ||
| (a) | p | r | q | s | |
| (b) | p | r | s | q | |
| (c) | r | p | r | s | |
| (d) | q | p | s | r | |
Correct Answer: C
Explanation:
$ v_{t} = +10 \mathrm{~ms}^{-1} $ and $ v_{f} = 0 $
$\therefore \quad \Delta v = v _ {f} - v _ {t} = - 1 0 \mathrm{ms} ^ {- 1}$
$a _ {\mathrm{av}} = \frac {\Delta v}{\Delta t} = \frac {- 1 0}{6} = - \frac {5}{3} \mathrm{ms} ^ {- 2}$
Total displacement = Area under v-t graph (with sign)
$= \frac {1}{2} \times 1 0 \times 2 - \frac {1}{2} \times 2 \times 1 0 - \frac {1}{2} \times 2 \times 1 0 = - 1 0 \mathrm{m}$
and acceleration = slope of v-t graph
$= - \frac {1 0}{2} = - 5 \mathrm{ms} ^ {- 2}$
Hence, A → r, B → p, C → r, D → s.
$\therefore \quad \Delta v = v _ {f} - v _ {t} = - 1 0 \mathrm{ms} ^ {- 1}$
$a _ {\mathrm{av}} = \frac {\Delta v}{\Delta t} = \frac {- 1 0}{6} = - \frac {5}{3} \mathrm{ms} ^ {- 2}$
Total displacement = Area under v-t graph (with sign)
$= \frac {1}{2} \times 1 0 \times 2 - \frac {1}{2} \times 2 \times 1 0 - \frac {1}{2} \times 2 \times 1 0 = - 1 0 \mathrm{m}$
and acceleration = slope of v-t graph
$= - \frac {1 0}{2} = - 5 \mathrm{ms} ^ {- 2}$
Hence, A → r, B → p, C → r, D → s.
Q4
Objective
Match the columns
Numerical
Let us call a motion, $A$ when velocity is positive and increasing, $A^{-1}$ when velocity is negative and increasing. $R$ when velocity is positive and decreasing and $R^{-1}$ when velocity is negative and decreasing. Now, match the following two columns for the given $s - t$ graph and mark the correct option from the codes given below.
| Column I | Column II | ||||||||
| (A) | M | (p) | ${A}^{-1}$ | ||||||
| (B) | N | (q) | ${R}^{-1}$ | ||||||
| (C) | P | (r) | A | ||||||
| (D) | Q | (s) | R | ||||||
| Codes | |||||||||
| A | B | C | D | A | B | C | D | ||
| (a) | p | r | q | s | (b) | r | s | p | q |
| (c) | s | p | r | q | (d) | q | p | s | r |
Correct Answer: B
Explanation:
For M, slope of s-t graph is positive and increasing.
Therefore, velocity of the particle is positive and increasing. Hence, it is A type motion. Similarly, N, P and Q can be observed from the slope.
Hence, A → r, B → s, C → p, D → q.
Therefore, velocity of the particle is positive and increasing. Hence, it is A type motion. Similarly, N, P and Q can be observed from the slope.
Hence, A → r, B → s, C → p, D → q.