Q576
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Figure shows the graph of the x-coordinate of a particle going along the x-axis as function of time.
The speed of particle at $t = 12.5 \, \text{s}$ (in ms $^{-1}$ ) is ......
The speed of particle at $t = 12.5 \, \text{s}$ (in ms $^{-1}$ ) is ......
Correct Answer: 2
Explanation:
At B, velocity is $2 \, ms^{-1}$ Slope of line AB is $v = -2 \, ms^{-1}$ So, speed of particle at t = 12.5 s is $2 \, ms^{-1}$
Q577
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle moving in a straight line covers half the distance with speed of $3 \, ms^{-1}$ . The other half of the distance is covered in two equal time intervals with a speeds of $4.5 \, ms^{-1}$ and $7.5 \, ms^{-1}$ , respectively. Find the average speed of the particle during this motion.
Correct Answer: 4
Explanation:
$v_{avg} = \frac{2v_{0}(v_{1} + v_{2})}{2v_{0} + v_{1} + v_{2}}$
$\Rightarrow v _ {a v g} = \frac {2 \times 3 (4 . 5 + 7 . 5)}{6 + 4 . 5 + 7 . 5} \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {a v g} = \frac {6 \times 1 2}{1 8} \mathrm{ms} ^ {- 1} = 4 \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {a v g} = \frac {2 \times 3 (4 . 5 + 7 . 5)}{6 + 4 . 5 + 7 . 5} \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {a v g} = \frac {6 \times 1 2}{1 8} \mathrm{ms} ^ {- 1} = 4 \mathrm{ms} ^ {- 1}$
Q578
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A body initially at rest moving along x-axis in such a way so that its acceleration displacement plot is as shown in figure. What will be the maximum velocity of particle in $ms^{-1}$ .
Correct Answer: 1
Explanation:
vdv = ads
$\Rightarrow \frac {v ^ {2}}{2} = \mathrm{Areaofa-sgraph}$
$\Rightarrow \frac {v ^ {2}}{2} = \frac {1}{2}$
$\Rightarrow v = 1 \mathrm{ms} ^ {- 1}$
$\Rightarrow \frac {v ^ {2}}{2} = \mathrm{Areaofa-sgraph}$
$\Rightarrow \frac {v ^ {2}}{2} = \frac {1}{2}$
$\Rightarrow v = 1 \mathrm{ms} ^ {- 1}$
Q579
Errorless
Graphical Questions
MCQ
An object is dropped from rest. Its $v - t$ graph is
A.
B.
C.
D.
E.
Q580
Errorless
Graphical Questions
MCQ
The motion of an v(m/sec) airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is ____ km
A.
9
B.
6
C.
3
D.
12
Q581
Errorless
JEE Section
MSQ
A particle moves with an initial velocity $v_{0}$ and retardation $\alpha v$ , where $v$ is its velocity at any time $t$
A.
The particle will cover a total distance $v_{0} / \alpha$
B.
The particle will come to rest after a time $1/\alpha$
C.
The particle will continue to move for a very long time
D.
The velocity of the particle will become $v_{0}/2$ after a time $1/\alpha$
Q582
Errorless
JEE Section
MCQ
Average velocity during the first 20 seconds
A.
$-10\mathrm{m / s}$
B.
$10~m / s$
C.
Zero
D.
None of these
Q583
Errorless
JEE Section
MCQ
Find the average acceleration during the first 20 seconds
A.
$\sqrt{3} m / s^2$
B.
$-\sqrt{3} m / s^2$
C.
Zero
D.
$1m / s^2$
Q584
Errorless
JEE Section
MCQ
During which interval is the motion retarded
A.
0 - 10
B.
10 - 20
C.
5 - 15
D.
None of these
Q585
Errorless
Matrix Match Type Questions
Numerical
A particle moves along a straight line such that its displacement s varies with time t as $s = \alpha + \beta t + \gamma t ^ {2}$
Match the following Column I with Column-II:
Match the following Column I with Column-II:
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → q; B → p; C → s; D → r
Explanation:
$\mathbf{A} \rightarrow \mathbf{q}; \mathbf{B} \rightarrow \mathbf{p}; \mathbf{C} \rightarrow \mathbf{s}; \mathbf{D} \rightarrow \mathbf{r}$ $s = \alpha + \beta t + \gamma t^2$ Velocity $v = \frac{ds}{dt} = \beta + 2\gamma t$ ;
Acceleration $a = \frac{d^2 s}{dt^2} = 2\gamma$ Thus particle moves with constant acceleration $2\gamma$ . $\therefore (A) \rightarrow q$ At $t = 1s, v = \beta + 2\gamma \therefore (C) \rightarrow s$ At $t = 0, s = \alpha$ $\therefore (D) \rightarrow r$ Average velocity during third second, $v_{av} = \frac{(\beta + 2\gamma t)_{t=2s} + (\beta + 2\gamma t)_{t=3s}}{2}$ $= \frac{\beta + 4\gamma + \beta + 6\gamma}{2} = \beta + 5\gamma \quad \therefore (B) \rightarrow p.$
Acceleration $a = \frac{d^2 s}{dt^2} = 2\gamma$ Thus particle moves with constant acceleration $2\gamma$ . $\therefore (A) \rightarrow q$ At $t = 1s, v = \beta + 2\gamma \therefore (C) \rightarrow s$ At $t = 0, s = \alpha$ $\therefore (D) \rightarrow r$ Average velocity during third second, $v_{av} = \frac{(\beta + 2\gamma t)_{t=2s} + (\beta + 2\gamma t)_{t=3s}}{2}$ $= \frac{\beta + 4\gamma + \beta + 6\gamma}{2} = \beta + 5\gamma \quad \therefore (B) \rightarrow p.$