Wondering what kind of questions that can be asked in next exam? Here you go!

Question.

A particle moving with an initial velocity

\hat{i} - 4 \hat{j} + 10 \hat{k}

has acceleration

\hat{i} + \hat{j} - 2 \hat{k}

. Its velocity at the end of 2 seconds, points along the unit vector:

\frac{1}{7} (- 3 \hat{i} + 2 \hat{j} - 6 \hat{k})

\frac{1}{3 8 2} (3 \hat{i} - 7 \hat{j} + 18 \hat{k})

\frac{1}{7} (3 \hat{i} - 2 \hat{j} + 6 \hat{k})

\frac{1}{7 7} (2 \hat{i} - 3 \hat{j} + 8 \hat{k})

Answer

Correct option is

$\frac{1}{7} (3 \hat{i} - 2 \hat{j} + 6 \hat{k})$

As we know
$v = u + a t$
$v = (\hat{i} - 4 \hat{j} + 10 \hat{k}) + (\hat{i} + \hat{j} - 2 \hat{k}) \times 2$
$v = (\hat{i} - 4 \hat{j} + 10 \hat{k}) + (2 \hat{i} + 2 \hat{j} - 4 \hat{k})$
$v = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$
Unit vector is given as,
$∣ v ∣ = 3^{2} + (- 2)^{2} + 6^{2}$
$∣ v ∣ = 49$
$∣ v ∣ = 7$
So, Velocity at the end of 2 sec, points along unit vector $= \frac{v}{∣ v ∣}$
$= \frac{3 i ^ - 2 j ^ + 6 k ^}{7}$
$= \frac{1}{7} (3 \hat{i} - 2 \hat{j} + 6 \hat{k})$

Question.

A particle moves in XY plane according to the law

x

a s i n (t)

and

y

a (1 - c o s (t))

where a is constant. The particle traces :

a parabola

a straight line equallyinclined to x and y axes

a circle

a distance proportional to time.

Answer

Correct option is

a parabola

$x = a sin t$
$y = a (1 - cos t)$
$\frac{d x}{d t} = a cos t$ $\frac{d y}{d t} = a sin t$
The particle is under the action and
two components
$V^{2} = (a cos t)^{2} + (a sin t)^{2} = a^{2} (sin^{2} t + cos^{2} t)$
$V = a$ constant.
Hence the particle trace is (A) parabola.

RELATED QESTIONS
Que - A particle moves in xy plane accordin to the law x=a sin w t and y=a (1- cos w t ) where a and w are costant . The particle traces.

a parabola

a straight line equally inclined to

x

and

y

axes

a circle

a distane proportional to time.

Answer

C::D

Question.

Which of the following displacement

(x)

vs. time

(t)

graphs is not possible?

Answer

Correct option is

A is not possible, because at a particular

time $t$ , displacement cannot have two values.

Can displacement be negative?

DISPLACEMENT: The term "displacement" refers to a shift in an object's location.

It's a vector quantity with a magnitude and a direction. ...

A vector can have values containing positive, negative as well as zero.

Therefore we can say that: Yes, the displacement can have negative values.

Question.

A man is going in a topless car with a velocity of $10.8 k m / h$ . It is raining vertically downwards. He has to hold the umbrella at an angle of $5 3^{o}$ to the vertical to protect himself from rain. The actual speed of the rain is $(c o s 5 3^{o} = \frac{3}{5})$

$2.25 m s^{- 1}$

$3.75 m s^{- 1}$

$0.75 m s^{- 1}$

$2.75 m s^{- 1}$

Answer

Correct option is

$2.25 m s^{- 1}$

Let the actual velocity of the rain = $x$
Then , relative to man, it is coming with $10.8 k m / h$ in horizontal and $x$ in vertical
$t a n 5 3^{o} = \frac{1 0 . 8}{x} = \frac{4}{3}$

$x = 3 \times \frac{1 0 . 8}{4} k m / h$

$3 \times \frac{1 0 . 8}{4} \times \frac{1}{3 . 6} = 2.25 m / s$

Related Question

A man is travelling at $10.8$ $km /h$ in a roofless car on a rainy day. He holds an umbrella at an angle $37$ $^{\circ}$ to the vertical to protect himself from the rain which is falling vertically downwards. What is the velocity of the rain?

(Given:

cos 37^{\circ} = \frac{4}{5}

)

$3 m/s$
$4 m/s$
$7 m/s$
$9 m/s$

Solution

The correct option is B $4 m/s$
Let $_{r}$ be the velocity of rain and $_{m}$ be the velocity of man travelling in the car.

Given: $cos 37^{\circ} = \frac{4}{5}$ , $v_{m} = 10.8 km/h,$
$sin 37^{\circ} = \frac{3}{5}$
So as, $tan 37^{\circ} = \frac{3}{4}$ .....(i)

From fig., $tan 37^{\circ} = \frac{v_{m}}{v_{r}}$ ......(ii)
on comparing (i) and (ii)
$\frac{v_{m}}{v_{r}} = \frac{3}{4}$
$\frac{10.8}{v_{r}} = \frac{3}{4}$
$v_{r} = 10.8 \times \frac{4}{3} = 14.4 km/h$

Converting $km/h$ into $m/s$ ,
$v_{r} = 14.4 \times \frac{5}{18}$
$= 4 m/s$

Question.

Consider the following statements A and B and identify the correct answer.

A) The speed acquired by a body when falling in a vacuum for a given time is dependent on the mass of the falling body.

B) A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for

5 s

. [

g = 10 m s^{- 2}

]

Both A & B are true

A is true but B is false

B is true but A is false

Both A & B are false

Answer

Question.

Consider the following statements A and B and identify the correct answer.

A) The speed acquired by a body when falling in a vacuum for a given time is dependent on the mass of the falling body.

5 s

. [

g = 10 m s^{- 2}

]

Both A & B are true

A is true but B is false

B is true but A is false

Both A & B are false

Answer

Correct option is

B is true but A is false

A) The Kinematic equation: $v = u + a t$ shows that the speed of the freely falling body depends on initial velocity u, acceleration a and time t. It does not depend on the mass. Therefore, statement A is false.

B) In general the distance traveled by the stone in some $n^{t h}$ second is given by:
$S_{n} = u + \frac{1}{2} a (2 n - 1)$
The stone falls freely from rest $∴ u = 0$
Let $n$ be the last second (time stone remains in air). Now distance traveled is iven by:
$S_{n} = \frac{1}{2} a (2 n - 1)$
This is also equal to the distance covered in first three seconds. This implies:
$\frac{1}{2} a (2 n - 1) = u t + \frac{1}{2} a t^{2} = \frac{1}{2} 3^{2} a = \frac{9}{2} a$
$\Rightarrow 2 n - 1 = 9$
$\Rightarrow n = 5$
The stone remains in air for 5 sec and hence statement B is true.

RELATED QUESTION-

A ball is dropped from a height of $19.6 m$ . The distance covered by it in the last second is

$19.6$

$14.7 m$

$4.8 m$

$9.8 m$

Correct option is

$14.7 m$

Let x be the distance covered in the last second. Now using $v = u + g t$ , we get
$2 g h = 2 g (h - x) + g 2 g (19.6) = 2 g (19.6 - x) + g$
Solving the above equation we get,
$x = 14.7$ m

Question.

A body starts from rest and moves with an uniform acceleration. The ratio of distance covered in the

n^{t h}

second to the distance covered in

n

seconds is :

(\frac{2}{n} - \frac{1}{n ^{2}})

(\frac{1}{n ^{2}} - \frac{1}{n})

(\frac{2}{n ^{2}} - \frac{1}{n})

(\frac{2}{n} + \frac{1}{n ^{2}})

Answer

Correct option is

$(\frac{2}{n} - \frac{1}{n ^{2}})$

Distance covered in $n$ seconds,
o $\to$ start from rest
$s = u t + \frac{1}{2} a t^{2}$
$s = \frac{1}{2} a n^{2} - - - - - (1)$

Distance covered in $n^{t h}$ second $= s_{n} - s_{n - 1}$

$\Rightarrow s_{n} - s_{n - 1} = \frac{1}{2} a n^{2} - \frac{1}{2} a (n - 1)^{2} - - - - - (2)$

So, $\frac{e q ^{n} ( 2 )}{e q ^{n} ( 1 )} = \frac{\frac{1}{2} a n ^{2} - \frac{1}{2} a ( n - 1 ) ^{2}}{\frac{1}{2} a n ^{2}} = \frac{2 n - 1}{n ^{2}}$ $= \frac{2}{n} - \frac{1}{n ^{2}}$

Question.

Raindrops are falling vertically with a velocity

10 m / s

. To a cyclist moving on a straight road the raindrops appear to be coming with a velocity of

20 m / s

. The velocity of cyclist is

10 m / s

103

m/s

20 m / s

203 m / s

Answer

Correct option is

$103$ m/s

Let the velocity of the cyclist be $v_{x}$
Velocity of the rain is $v_{y} = 10 m / s$
Relative velocity $= - v_{x}$ along $x$ and $v_{y}$ along $y$
Hence relative velocity $= v_{x}^{2} + v_{y}^{2} = 20 m s^{- 1}$
$v_{x}^{2} + 100 = 400$
$v_{x}^{2} = 400 - 100 = 300$
Therefore, the velocity of cyclist, $v_{x} = 103 m s^{- 1}$
Hence option $B$ is correct.

Question.

A body is thrown vertically upwards. Which one of the following graphs correctly represents the velocity vs time?

Answer

Correct option is

Velocity of a body thrown vertically upwards can be given as:
$v_{f} = v_{i} + a t$
where $a = - g$ (always acts in downward direction)
$∴ v_{f} = v_{i} - g t$
for $t < \frac{v _{i}}{g}$ , $v_{f}$ is positive
at $t = \frac{v _{i}}{g}$ , we get $v_{f} = 0$
for $t > \frac{v _{i}}{g}$ , $v_{f}$ is negative.
also the slope of $v$ vs $t$ graph is $- g$ [Negative]
From the above explanations, we can conclude that the velocity starts from a positive value, starts deceasing, reaches zero and keeps on decreasing for be a negative value.
Hence, the correct option is $(D)$

Question.

A ball is dropped vertically from height

h

and is bouncing elastically on the floor (see figure). Which of the following plots best depicts the acceleration of the ball as a function of time.

Answer

Correct option is

During free fall of the ball, gravitational force acts on the body in downward direction always and so the acceleration of the ball remains in vertically downward direction which is constant with time. Its acceleration becomes positive only at single instant of time during the collision of ball with the ground. The ground imparts an upward impulse and hence the upward acceleration on the ball which makes the ball to move upward, but as soon as the ball looses contact with the ground, its accelerates in the downward direction again. Hence, option B is correct.

Question.

Position (Km) - Time (min.) graph is as shown for two cars

A

and

B

. Both collide at time

t = 150

minute. Then the distance of position

R

of accident from the starting point

Q

of car

A

will be: (Initial distance between the two cars is

500

km) (Position in the graph shows the distance of the two cars from the point

Q

)

200 k m

300 k m

250 k m

400 k m

Answer

Correct option is

$300 k m$

We need to calculate the distance of position 'R' of accident from the starting point 'Q' of car A i.e. the y coordinate of point R in the graph.
It is given that the angle of PR with the y-axis is $3 7^{o}$
Thus angle of PR with x-axis will be $12 7^{o}$
Thus slope of line PR is $t a n (12 7^{o}) = - \frac{4}{3}$
Equation of line will be
$y = \frac{- 4}{3} x + 500 \Rightarrow y = \frac{- 4}{3} \times 150 + 500 \Rightarrow y = 300 k m$ (x=150 min)

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CLASS 11 | CHAPTER 3 | Motion In A Straight Line-Types Of Questions | PHYSICS | NEET

$\frac{1}{7} (3 \hat{i} - 2 \hat{j} + 6 \hat{k})$

a parabola

$x = a sin t$
$y = a (1 - cos t)$
$\frac{d x}{d t} = a cos t$ $\frac{d y}{d t} = a sin t$
The particle is under the action and
two components
$V^{2} = (a cos t)^{2} + (a sin t)^{2} = a^{2} (sin^{2} t + cos^{2} t)$
$V = a$ constant.
Hence the particle trace is (A) parabola.

RELATED QESTIONS
Que - A particle moves in xy plane accordin to the law x=a sin w t and y=a (1- cos w t ) where a and w are costant . The particle traces.

A is not possible, because at a particular

time $t$ , displacement cannot have two values.

Can displacement be negative?

DISPLACEMENT: The term "displacement" refers to a shift in an object's location.

It's a vector quantity with a magnitude and a direction. ...

A vector can have values containing positive, negative as well as zero.

Therefore we can say that: Yes, the displacement can have negative values.

A man is going in a topless car with a velocity of $10.8 k m / h$ . It is raining vertically downwards. He has to hold the umbrella at an angle of $5 3^{o}$ to the vertical to protect himself from rain. The actual speed of the rain is $(c o s 5 3^{o} = \frac{3}{5})$

$2.25 m s^{- 1}$

$3.75 m s^{- 1}$

$0.75 m s^{- 1}$

$2.75 m s^{- 1}$

$2.25 m s^{- 1}$

Let the actual velocity of the rain = $x$
Then , relative to man, it is coming with $10.8 k m / h$ in horizontal and $x$ in vertical
$t a n 5 3^{o} = \frac{1 0 . 8}{x} = \frac{4}{3}$

$x = 3 \times \frac{1 0 . 8}{4} k m / h$

$3 \times \frac{1 0 . 8}{4} \times \frac{1}{3 . 6} = 2.25 m / s$

A man is travelling at $10.8$ $km /h$ in a roofless car on a rainy day. He holds an umbrella at an angle $37$ $^{\circ}$ to the vertical to protect himself from the rain which is falling vertically downwards. What is the velocity of the rain?

B is true but A is false

A ball is dropped from a height of $19.6 m$ . The distance covered by it in the last second is

$19.6$

$14.7 m$

$4.8 m$

$9.8 m$

$14.7 m$

Let x be the distance covered in the last second. Now using $v = u + g t$ , we get
$2 g h = 2 g (h - x) + g 2 g (19.6) = 2 g (19.6 - x) + g$
Solving the above equation we get,
$x = 14.7$ m

$(\frac{2}{n} - \frac{1}{n ^{2}})$

$103$ m/s

$300 k m$

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CLASS 11 | CHAPTER 3 | Motion In A Straight Line-Types Of Questions | PHYSICS | NEET

71​(3i^−2j^​+6k^)

a parabola

x=asinty=a(1−cost)dtdx​=acostdtdy​=asintThe particle is under the action and two components V2=(acost)2+(asint)2=a2(sin2t+cos2t)V=a constant.Hence the particle trace is (A) parabola.

RELATED QESTIONSQue - A particle moves in xy plane accordin to the law x=a sin w t and y=a (1- cos w t ) where a and w are costant . The particle traces.

A is not possible, because at a particular

time t, displacement cannot have two values.

Can displacement be negative?

DISPLACEMENT: The term "displacement" refers to a shift in an object's location.

It's a vector quantity with a magnitude and a direction. ...

A vector can have values containing positive, negative as well as zero.

Therefore we can say that: Yes, the displacement can have negative values.

A man is going in a topless car with a velocity of 10.8km/h. It is raining vertically downwards. He has to hold the umbrella at an angle of 53o to the vertical to protect himself from rain. The actual speed of the rain is (cos53o=53​)

2.25 ms−1

3.75 ms−1

0.75 ms−1

2.75 ms−1

2.25 ms−1

Let the actual velocity of the rain = xThen , relative to man, it is coming with 10.8km/h in horizontal and x in verticaltan53o=x10.8​=34​ x=3×410.8​km/h3×410.8​×3.61​=2.25m/s

A man is travelling at 10.8 km /h in a roofless car on a rainy day. He holds an umbrella at an angle 37∘ to the vertical to protect himself from the rain which is falling vertically downwards. What is the velocity of the rain?

B is true but A is false

A ball is dropped from a height of 19.6m. The distance covered by it in the last second is

19.6

14.7m

4.8m

9.8m

14.7m

Let x be the distance covered in the last second. Now using v=u+gt, we get 2gh​=2g(h−x)​+g2g(19.6)​=2g(19.6−x)​+gSolving the above equation we get,x=14.7m

(n2​−n21​)

Distance covered in n seconds,o → start from rests=ut+21​at2s=21​an2−−−−−(1)Distance covered in nth second =sn​−sn−1​⇒sn​−sn−1​=21​an2−21​a(n−1)2−−−−−(2)So, eqn(1)eqn(2)​=21​an221​an2−21​a(n−1)2​=n22n−1​=n2​−n21​

103 m/s

Let the velocity of the cyclist be vx​Velocity of the rain is vy​=10m/sRelative velocity =−vx​ along x and vy​ along yHence relative velocity =vx2​+vy2​​=20ms−1vx2​+100=400vx2​=400−100=300Therefore, the velocity of cyclist, vx​=103​ms−1Hence option B is correct.

300km

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$\frac{1}{7} (3 \hat{i} - 2 \hat{j} + 6 \hat{k})$

$x = a sin t$
$y = a (1 - cos t)$
$\frac{d x}{d t} = a cos t$ $\frac{d y}{d t} = a sin t$
The particle is under the action and
two components
$V^{2} = (a cos t)^{2} + (a sin t)^{2} = a^{2} (sin^{2} t + cos^{2} t)$
$V = a$ constant.
Hence the particle trace is (A) parabola.

RELATED QESTIONS
Que - A particle moves in xy plane accordin to the law x=a sin w t and y=a (1- cos w t ) where a and w are costant . The particle traces.

time $t$ , displacement cannot have two values.

A man is going in a topless car with a velocity of $10.8 k m / h$ . It is raining vertically downwards. He has to hold the umbrella at an angle of $5 3^{o}$ to the vertical to protect himself from rain. The actual speed of the rain is $(c o s 5 3^{o} = \frac{3}{5})$

$2.25 m s^{- 1}$

$3.75 m s^{- 1}$

$0.75 m s^{- 1}$

$2.75 m s^{- 1}$

$2.25 m s^{- 1}$

Let the actual velocity of the rain = $x$
Then , relative to man, it is coming with $10.8 k m / h$ in horizontal and $x$ in vertical
$t a n 5 3^{o} = \frac{1 0 . 8}{x} = \frac{4}{3}$

$x = 3 \times \frac{1 0 . 8}{4} k m / h$

$3 \times \frac{1 0 . 8}{4} \times \frac{1}{3 . 6} = 2.25 m / s$

A man is travelling at $10.8$ $km /h$ in a roofless car on a rainy day. He holds an umbrella at an angle $37$ $^{\circ}$ to the vertical to protect himself from the rain which is falling vertically downwards. What is the velocity of the rain?

A ball is dropped from a height of $19.6 m$ . The distance covered by it in the last second is

$19.6$

$14.7 m$

$4.8 m$

$9.8 m$

$14.7 m$

Let x be the distance covered in the last second. Now using $v = u + g t$ , we get
$2 g h = 2 g (h - x) + g 2 g (19.6) = 2 g (19.6 - x) + g$
Solving the above equation we get,
$x = 14.7$ m

$(\frac{2}{n} - \frac{1}{n ^{2}})$

$103$ m/s

$300 k m$