Motion in A Straight Line Test

Result

Motion in A Straight Line Test - 11

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is

A
4.0 m/s

B
5.0 m/s

C
5.5 m/s

D
4.8 m/s

Solution

If \(t_1\) and 2\(t_2\) are the time taken by particle to cover first and second half distance, respectively, then

\(t_1 = \frac{\frac{x}{2}}{3}\) = \(\frac{x}{6}\)

\( x_1\)= 4.5 \(t_2\) and \(x_2\) = 7.5\(t_2\)

So, \(x_1 + x_2 = \frac{x}{2}\)

⇒ 4.5 \(t_2\) + 7.5\(t_2\) = \(\frac{x}{2}\)

\(t_2 = -\frac{x}{24}\)

Total time \(t_1 + 2t_2\)

= \(\frac{x}{6} + \frac{x}{12} = \frac{x}{4}\)

So, average speed 4 m/sec.
Question 2
1 / -0

A body is slipping from an inclined plane of height h and length l. If the angle of inclination is θ, the time taken by the body to come from the top to the bottom of this inclined plane is

A
\(\sqrt{\frac{2h}{g}}\)

B
\(\sqrt{\frac{2l}{g}}\)

C
\(\frac{1}{sin \theta}\)\(\sqrt{\frac{2h}{g}}\)

D
\(sin \theta\)\(\sqrt{\frac{2h}{g}}\)

Solution

Force down the plane = mg sinθ, Acceleration down the plane = g sinθ since

\(l = 0 + \frac{1}{2}g sin \;\theta t^2\)

\(t^2 = \frac{2l}{g \; sin \theta} = \frac{2h}{g sin^2 \theta}\)

⇒ t = \(\frac{1}{sin \theta} \sqrt{\frac{2h}{g}}\)
Question 3
1 / -0

A particle moves with uniform acceleration along a straight line AB. Its velocities at A and B are 2 m/s and 14 m/s, respectively. M is the mid-point of AB. The particle takes \(t_1\) seconds to go from A to M and \(t_2\) seconds to go from M to B. Then \(\frac{t_2}{t_1}\) is

A
1:1

B
2:1

C
1:2

D
3:1

Solution

\(v^2 = 2^2 + \frac{2as}{2}\)

⇒ \(v^2 - 4 = as \)

\(14^2 - v^2 = \frac{2as}{2}\)

⇒ 196 - \(v^2 = as\)

From (i) and (ii)

\(v^2 - 4 = 196 - v^2\)

⇒ v = 10 m/s

Now,

\(t_1 = \frac{v - 2}{a} = \frac{10 - 2}{a} = \frac{8}{a}\)

\(t_2 = \frac{14 - v}{a} = \frac{14 - 10}{a} = \frac{4}{a}\)

⇒ \(\frac{t_2}{t_1} = \frac{\frac{4}{a}}{\frac{8}{a}} = \frac{1}{2}\)
Question 4
1 / -0

A police party is chasing a dacoit in a jeep which is moving at a constant speed v. The dacoit is on a motorcycle. When he is at a distance x from the jeep, he accelerates from rest at a constant rate. Which of the following relations is true if the police are able to catch the dacoit?

A
\(v^2 \leq \alpha x\)

B
\(v^2 \leq 2 \alpha x\)

C
\(v^2 \geq 2 \alpha x\)

D
\(v^2 \geq \alpha x\)

Solution

If police is able to catch the dacoit after time t, then

vt = x + \(\frac{1}{2}at^2\)

This gives \(\frac{a}{2}t^2 \) - vt + x = 0

Or t = \(\frac{v \pm \sqrt{v^2 - 2ax}}{a}\)

For t to be real, \(v^2 \geq 2 \alpha x\)
Question 5
1 / -0

A body starts from rest and travels ‘s’ m in 2nd second, then acceleration is

A
\((2s) m/s^2\)

B
\((3s)m/s^2\)

C
\((\frac{2}{3}s) m/s^2\)

D
\((\frac{3}{2}s)m/s^2\)

Solution

\(s_n = u + \frac{a}{2}(2n - 1)\)

or, \(s = \frac{a}{2}\)(2 x 2 - 1)

⇒ a = \(\frac{2}{3}sm/s^2\)
Question 6
1 / -0

The displacement x of a particle at the instant when its velocity is v is given by v = \(\sqrt{3x + 16}\). Its acceleration and initial velocity are

A
1.5 units, 4 units

B
3 units, 4 units

C
16 units, 1.6 units

D
16 units, 3 units

Solution

\(v = \sqrt{3x + 16}\)

⇒ \(v^2\) = 3x + 16

⇒ \(v^2\) - 16 = 3x

Comparing with \(v^2 - u^2 \) = 2aS,

we get, u = 4 units, 2a = 3

or a = 1.5 units
Question 7
1 / -0

A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance \(s_1 \) in the first 10 seconds and distance \(s_2\) in the next 10 seconds, then

A
\(s_2 = s_1\)

B
\(s_2 = 2s_1\)

C
\(s_2 = 3s_1\)

D
\(s_2 = 4s_1\)

Solution

Let a be the constant acceleration of the particle. Then

\(s = ut + \frac{1}{2}at^2\)

or \(s_1 = 0 + \frac{1}{2} \times a \times (10)^2 \) = 50a

and \(s_2 = \Big[0 + \frac{1}{2}a(20)^2 \Big]\) - 50a = 150 a

\(\therefore s_2 = 3s_1\)
Question 8
1 / -0

If distance covered by a particle is zero, what can you say about its displacement?

A
It may or may not be zero

B
It cannot be zero

C
It is negative

D
It must be zero

Solution

Distance covered by a particle is zero only when it is at rest. Therefore, its displacement must be zero.
Question 9
1 / -0

The displacement-time graphs of two particles A and B are straight lines making angles of 30° and 60° respectively with the time axis. If the velocity of A is \(v_A\) and that of B is \(v_B\), the value of \(\frac{v_A}{v_B}\) is

A
\(\frac{1}{2}\)

B
\(\frac{1}{\sqrt 3}\)

C
\(\sqrt3\)

D
\(\frac{1}{3}\)

Solution

\(v_A\) = tan30° and \(v_B\) = tan60°

\(\therefore \frac{V_A}{V_B} = \frac{tan 30^o}{tan 60^o}\)

= \(\frac{\frac{1}{\sqrt3}}{\sqrt3}\) = \(\frac{1}{3}\)
Question 10
1 / -0

A particle moves in straight line with velocity 6 m/s and 3 m/s for time intervals which are in ratio 1:2. Find average velocity.

A
2m/s

B
3m/s

C
4m/s

D
3m/s

Solution

Let the times be x and 2x seconds.

Then the distances travelled will be 6x and 6x meters

Average velocity = total distance travelled / total time taken

therefore, avg vel= 12x/3x

Average velocity = 4m/s
Question 11
1 / -0

A body covers 26, 28, 30, 32 meters in 10th, 11th, 12th and 13th seconds respectively. The body

A
From rest and moves with uniform velocity

B
From rest and moves with uniform acceleration

C
With an initial velocity and moves with uniform acceleration

D
With an initial velocity and moves with uniform velocity

Solution

The distance covered in nth second is where u is initial velocity & a is acceleration

From eqn. (i) and (ii)

\(S_n = u + \frac{1}{2}(2n - 1)a\)

26 = u + \(\frac{19a}{2}\) ....(i)

28 = u + \(\frac{21a}{2}\) .....(ii)

\(30 = u +\frac{23a}{2}\) .....(iii)

\(32 = u + \frac{25a}{2}\) .....(iv)

We get u = 7m/sec, a = 2\(m/sec^2\)

\(\therefore\) The body starts with initial velocity u = 7m/sec and moves with uniform acceleration a = 2\(m/sec^2\).
Question 12
1 / -0

A car moves with a speed of 60 km/hr from point A to point B and then with the speed of 40 km/hr from point B to point C. Further, it moves to a point D with a speed equal to its average speed between A and C. Points A, B, C and D are collinear and equidistant. The average speed of the car between A and D is

A
30 km/hr

B
50 km/hr

C
48 km/hr

D
60 km/hr

Solution

Let the points A, B, C and D be separated by 1 km. Then

\(t_{AB} = \frac{1}{60} hr,\) \(t_{BC} = \frac{1}{40}hr\)

\(\therefore < v_{AC} > = \frac{1 + 1}{\frac{1}{60} + \frac{1}{40}}\) = 48 km/hr

⇒ \(t_{CD}= \frac{1}{48} hr\)

Now < \(v_{AD} > = \frac{1 + 1 + 1}{\frac{1}{60} + \frac{1}{40} + \frac{1}{48}}\)

= 48 km/hr
Question 13
1 / -0

A bus starts moving with acceleration 2 \(m/s^2\). A cyclist 96 m behind the bus starts simultaneously towards the bus at 20 m/s. After what time will he be able to overtake the bus?

A
4 sec

B
8 sec

C
18 sec

D
16 sec

Solution

Velocity of bus \(V_b = 0,\) Velocity of cyclist \(V_c\) = 20m/s,

Acceleration of bus \(a_b=2m/s^2\), Acceleration of cyclist \(a_c = 0\)

Relative velocity \(V_{cb}=V_c-V_b\) = 20m/s

Relative acceleration \(a_{cb}=a_c-a_b=-2m/s^2\)

Relative separation = 96m

Using \(s=ut+\frac{1}{2}at^2\)

\(\Rightarrow\) \(96 = 20t -\frac{1}{2}2t^2\)

\(\Rightarrow t^2-20t+96=0\)

Solving the equation, we get t = 8s and t = 12s

Hence, after 8s, cyclist will overtake the bus.
Question 14
1 / -0

If a train travelling at 20 m/s is to be brought to rest in a distance of 200 m, then its retardation should be

A
\(1 m/s^2\)

B
\(2 m/s^2\)

C
\(10 m/s^2\)

D
\(20 m/s^2\)

Solution

\(v^2 - u^2 = 2aS\)

⇒ a = \(\frac{u^2}{2aS}\)

= \(\frac{(20)^2}{2 \times 200}\)

= \(1 m/s^2\)
Question 15
1 / -0

A bike accelerates from rest at a constant rate 5\(m/s^2\) for some time after which it decelerates at a constant rate 3 \(m/s^2\) to come to rest. If the total time elapsed is 8 second, the maximum velocity acquired by the bike is given by.

A
5 m/s

B
10 m/s

C
12 m/s

D
15 m/s

Solution

\(V_{max} = 0 + 5t_1\) ....(i)

0 = \(V_{max} - 3t_2 \) ....(ii)

⇒ \(t_1 + t_2 = 8 = V_{max}(\frac{1}{5} + \frac{1}{3})\)

⇒ \(V_{max} = \frac{8 \times 15}{8}\)

= 15 m/s
Question 16
1 / -0

Two particles P and Q simultaneously start moving from point A with velocities 15 m/s and 20 m/s respectively. The two particles move with acceleration equal in magnitude but opposite in direction. When P overtakes Q at B then its velocity is 30 m/s. The velocity of Q at point B will be

A
30 m/s

B
5 m/s

C
20 m/s

D
15 m/s

Solution

As magnitude of acceleration is same

\(\frac{30 - 15}{t} = \frac{20 - V_Q}{t}\) 15 = 20 - \(V_Q\) ⇒ 5 m/s
Question 17
1 / -0

A man throws balls with same speed vertically upwards one after the other at an interval of 2 sec. What should be the speed of throw so that more man two balls are in air at any time?

A
Only with speed 19.6 m/s

B
More than 19.6 m/s

C
At least 9.8 m/s

D
Any speed less than 19.6 m/s

Solution

Height attained by balls in 2 sec is

= \(\frac{1}{2}\) x 9.8 x 4

= 19.6 m

The same distance will be covered in 2 second (for descent) Time interval of throwing balls, remaining same. So, for two balls remaining in air, the time of ascent or descent must be greater than 2 seconds. Hence speed of balls must be greater than 19.6 m/sec.
Question 18
1 / -0

A ball dropped from a point A falls down vertically to C, through the midpoint B. The descending time from A to B and that from A to C are in the ratio

A
1:1

B
1:2

C
1:3

D
1:\(\sqrt2\)

Solution

For A to B

S = \(\frac{1}{2}gt^2\) .....(i)

For A to C S' = \(\frac{1}{2}gt'^2\) ....(ii)

Dividing (i) by (ii)

we get \(\frac{t}{t'} = \frac{1}{\sqrt 2}\)
Question 19
1 / -0

A stone falls freely under gravity. It covers distances \(h_1, h_2\) and \(h_3\) in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between \(h_1, h_2\) and \(h_3\) is

A
\(h_1 = \frac{h_2}{3} = \frac{h_3}{5}\)

B
\(h_2 = 3h_1 \) and \(h_3 = 3h_2\)

C
\(h_1 = h_2 = h_3\)

D
\(h_1 = 2h_2 = 3h_3\)

Solution

\(\therefore h = \frac{1}{2}gt^2\)

\(\therefore h_1 = \frac{1}{2}g(5)^2 = 125\)

\(h_1 + h_2\)= \(\frac{1}{2}g(10)^2 = 500\)

⇒ \(h_2 = 375\)

\(\therefore h_1 + h_2 + h_3\)

= \(\frac{1}{2}g(15)^2 = 1125\)

⇒ \(h_3 =625\)

\( h_2 = 3h_1, h_3 = 5h_1\)

or \(h_1 = \frac{h_2}{3} = \frac{h_3}{5}\)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 11

Result

Motion in A Straight Line Test - 11

Score

-

Rank

-

SHARING IS CARING

Question 1

4.0 m/s

5.0 m/s

5.5 m/s

4.8 m/s

Solution

Question 2

\(\sqrt{\frac{2h}{g}}\)

\(\sqrt{\frac{2l}{g}}\)

\(\frac{1}{sin \theta}\)\(\sqrt{\frac{2h}{g}}\)

\(sin \theta\)\(\sqrt{\frac{2h}{g}}\)

Solution

Question 3

1:1

2:1

1:2

3:1

Solution

Question 4

\(v^2 \leq \alpha x\)

\(v^2 \leq 2 \alpha x\)

\(v^2 \geq 2 \alpha x\)

\(v^2 \geq \alpha x\)

Solution

Question 5

\((2s) m/s^2\)

\((3s)m/s^2\)

\((\frac{2}{3}s) m/s^2\)

\((\frac{3}{2}s)m/s^2\)

Solution

Question 6

1.5 units, 4 units

3 units, 4 units

16 units, 1.6 units

16 units, 3 units

Solution

Question 7

\(s_2 = s_1\)

\(s_2 = 2s_1\)

\(s_2 = 3s_1\)

\(s_2 = 4s_1\)

Solution

Question 8

It may or may not be zero

It cannot be zero

It is negative

It must be zero

Solution

Question 9

\(\frac{1}{2}\)

\(\frac{1}{\sqrt 3}\)

\(\sqrt3\)

\(\frac{1}{3}\)

Solution

Question 10

2m/s

3m/s

4m/s

3m/s

Solution

Question 11

From rest and moves with uniform velocity

From rest and moves with uniform acceleration

With an initial velocity and moves with uniform acceleration

With an initial velocity and moves with uniform velocity

Solution

Question 12

30 km/hr

50 km/hr

48 km/hr

60 km/hr

Solution