Motion in A Straight Line Test

Result

Motion in A Straight Line Test - 43

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 car is moving along a straight road with a uniform acceleration. It passes through two points $$P$$ and $$Q$$ separated by a distance, with velocity $$30\ km/hr$$ and $$40\ km/hr$$ respectively. The velocity of the car midway between $$P$$ and $$Q$$ is

A
$$35.35\ km/hr$$

B
$$23.3\ km/hr$$

C
$$283\sqrt {2} km/hr$$

D
none of these.

Solution

using equations of motion $$v^2-u^2=2aS$$
$$v=40,u=30,a=constant,S=?$$
we get $$S=\dfrac{40^2-30^2}{2a}=>S=\dfrac{350}{a}$$
now given $$S_{middle}=\dfrac{1}{2}\times \dfrac{350}{a},a=constant,u=30,v=?$$
$$v_{middle}^2=2aS_{middle}+u^2=>v_{middle}^2=2a\times \dfrac{350}{2a}+30^2=>v_{middle}=\sqrt{350+900}$$
$$v_{middle}=35.35km/hr$$
Question 2
1 / -0

A body $$X$$ is projected upwards with a velocity of $$98\ ms^{-1}$$, after $$4s$$, a second body $$Y$$ is also projected upwards with the same $$Y$$ is also projected upwards with the same initial velocity. Two bodies will meet after

A
$$8\ s$$

B
$$10\ s$$

C
$$12\ s$$

D
$$14\ s$$

Solution

Let $$t$$ second be the time of flight of the first body after meeting, then $$(t - 4)$$ second will be the time of flight of the second body.
Since, $$h_{1} = h_{2}$$
$$\therefore 98t - \dfrac {1}{2}gt^{2} = 98 (t - 4)g (t - 4)^{2}$$
On solving, $$t = 12s$$.
Question 3
1 / -0

A man of mass $$60$$kg and a boy of mass $$30$$kg are standing together on frictionless ice surface. If they push each other apart man moves away with a speed of $$0.4$$m/s relative to ice. After $$5$$sec they will be away from each other at a distance of.

A
$$9.0$$m

B
$$3.0$$m

C
$$6.0$$m

D
$$30$$,

Solution

The man and the boy move in opposite directions.
Momentum of man$$=$$ momentum of boy
$$60\times 0.4=30\times v$$
or velocity of the boy $$v=0.8ms^{-1}$$
$$\therefore$$ Relative velocity $$=0.4+0.8=1.2ms^{-1}$$
$$\therefore$$ Distance between them in $$5$$sec $$ =1.2\times 5=6.0$$m.
Question 4
1 / -0

If the distance travel by a uniformly accelerated particle in $$pth, qt$$ and $$rth$$ second are $$a, b$$ and $$c$$ respectively. Then

A
$$(q - r) a + (r - p) b + (p - q) c = 1$$

B
$$(q - r) a + (r - p) b + (p - q) c = -1$$

C
$$(q - r) a + (r - p) b + (p - q) c = 0$$

D
$$(q + r) a + (r + p) b + (p + q) c = 0$$

Solution

Let $$u$$ be the initial velocity of the particle and $$f$$ be acceleration of the particle. Then
$$a = u + \dfrac {1}{2} f(2p - 1) ..... (i)$$
$$b = u + \dfrac {1}{2}f(2q - 1) .... (ii)$$
$$c = u + \dfrac {1}{2} f(2r - 1) ..... (iii)$$
Now, multiplying Eq. (i) by $$(q - r)$$, Eq. (ii) by $$(r - p)$$ and Eq. (iii) by $$(p - q)$$ and adding, we get $$a(q - r) + b(r - p) + c(p - q) = a(0) + \dfrac {1}{2} f(0)$$
$$\therefore a(q - r) + b(r - p) + c(p - q) = 0$$.
Question 5
1 / -0

If a coin is dropped in a lift it takes $$t_{1}$$ time to reach the floor and takes $$t_{2}$$ time when lift is moving up with constant acceleration, when which one of the following relation is correct?

A
$$t_{1} = t_{2}$$

B
$$t_{1} > t_{2}$$

C
$$t_{2} > t_{1}$$

D
$$t_{1} > > t_{2}$$

Solution

Time $$t_{1}$$ for stationary lift $$= \sqrt {\dfrac {2h}{g}}$$
When lift is moving up with constant acceleration, then
$$t_{2} = \sqrt {\dfrac {2h}{g + a}}$$
$$\therefore t_{1} > t_{2}$$.
Question 6
1 / -0

A body starts from rest acquires a velocity $$v$$ in time $$T$$. The work done on the body in time$$t$$ will be proportional to

A
$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ 2 } } \right) { t }^{ 2 }$$

B
$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ 2 } } \right) { t }$$

C
$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ } } \right) { t }^{ 2 }$$

D
$$\left( \cfrac { { V }^{ } }{ { T }^{ } } \right) { t }^{ 2 }$$

Solution

The acceleration of the body $$a=\cfrac { v }{ T } $$
Now, work done $$=\cfrac { 1 }{ 2 } m{ v }^{ 2 }\propto { v }^{ 2 }$$
Here $$v=u+at$$
so, work done $${ v }^{ 2 }\propto { a }^{ 2 }{ t }^{ 2 }$$
or $$W\propto \left( \cfrac { { v }^{ 2 } }{ { T }^{ 2 } } \right) { t }^{ 2 }\quad $$
Question 7
1 / -0

Initial speed of an alpha particle side a tube a length $$4\ m$$ is $$1\ km/s$$, if it is accelerated in the tube and comes out with a speed of $$9\ km/s$$, then the time for which the particle remains inside the tube is

A
$$8\times 10^{-3}s$$

B
$$8\times 10^{-4}s$$

C
$$80\times 10^{-3}s$$

D
$$800\times 10^{-3}s$$

Solution

Initial speed, $$u = 1\ km/s = 1000\ m/s$$
Final speed, $$v = 9\ km/s = 9000\ m/s$$
By using the relation
$$v^{2} = u^{2} + 2as$$
$$(9000)^{2} = (1000)^{2} + 2\times a \times 4$$
$$a = 10^{7} m/s^{2}$$
$$\therefore$$ the time for which the particle remains in the tube
$$v = u + at$$
or $$t = \dfrac {v - u}{a} = \dfrac {9000 - 1000}{10^{7}}$$
$$= 8\times 10^{-4}s$$.
Question 8
1 / -0

A ball thrown upward from the top of a tower with speed $$v$$ reaches the ground in $$t_{1}$$ second. If this ball is thrown downward from the top of the same tower with speed $$v$$ it reaches the ground in $$t_{2}$$ second. In what time the ball shall reach the ground if it is allowed to fall freely under gravity from the top of the tower?

A
$$\dfrac {t_{1} + t_{2}}{2}$$

B
$$\dfrac {t_{1} - t_{2}}{2}$$

C
$$\sqrt {t_{1}t_{2}}$$

D
$$t_{1} + t_{2}$$

Solution

$$h = -vt_{1} + \dfrac {1}{2} gt_{1}^{2}, h = vt_{2} + \dfrac {1}{2} gt_{2}^{2}$$
Now, $$\dfrac {h}{t_{1}} + \dfrac {h}{t_{2}} = \dfrac {1}{2}g(t_{1} + t_{2})$$ or $$h = \dfrac {1}{2} gt_{1}t_{2}$$
Also, $$h = \dfrac {1}{2}gt^{2} \therefore t^{2} = t_{1}t_{2}$$or $$t = \sqrt {t_{1}t_{2}}$$.
Question 9
1 / -0

The time required to stop a car of mass $$800\ kg$$ moving at a speed of $$20\ ms^{-1}$$ over a distance of $$25\ m$$ is

A
$$2\ s$$

B
$$2.5\ s$$

C
$$4\ s$$

D
$$4.5\ s$$

E
$$1\ s$$

Solution

Given, $$u = 0\ ms^{-1}, v = 20\ ms^{-1}$$
Mass of car $$= 800\ kg, s = 25\ m$$
We know that, distance covered,
$$s = \left (\dfrac {u + v}{2}\right ) t$$
$$\Rightarrow 25 = \left (\dfrac {0 + 20}{2}\right )t$$
$$= \dfrac {25}{10} = 2.5 s$$.
Question 10
1 / -0

The driver of an express train suddenly sees the red light signal 50 m ahead and applies the brakes .If the average deceleration during braking is $$10.0 ms^{-2}$$ and the reaction time of the driver is 0.75 sec , the minimum speed at which the train should be moving so as not to cross the red signal is

A
27 km/hr

B
144 km/hr

C
72 km/hr

D
83 km/hr

Solution

Let,
Displacement of Train in 0.75 sec will be
$$S_1 = ut = 0.75 \ u \ m$$
where, u - initial velocity of train.
$$v^2 = u^2 + 2aS$$
$$v = 0 ms^{-1}, S = 50- S_1$$
$$u^2 - 2 \times 10 \times(50 - 0.75u) = 0 $$
solving for u we get
$$u = 144 \ km/hr$$

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 - 43

Result

Motion in A Straight Line Test - 43

Score

-

Rank

-

SHARING IS CARING

Question 1

$$35.35\ km/hr$$

$$23.3\ km/hr$$

$$283\sqrt {2} km/hr$$

none of these.

Solution

Question 2

$$8\ s$$

$$10\ s$$

$$12\ s$$

$$14\ s$$

Solution

Question 3

$$9.0$$m

$$3.0$$m

$$6.0$$m

$$30$$,

Solution

Question 4

$$(q - r) a + (r - p) b + (p - q) c = 1$$

$$(q - r) a + (r - p) b + (p - q) c = -1$$

$$(q - r) a + (r - p) b + (p - q) c = 0$$

$$(q + r) a + (r + p) b + (p + q) c = 0$$

Solution

Question 5

$$t_{1} = t_{2}$$

$$t_{1} > t_{2}$$

$$t_{2} > t_{1}$$

$$t_{1} > > t_{2}$$

Solution

Question 6

$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ 2 } } \right) { t }^{ 2 }$$

$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ 2 } } \right) { t }$$

$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ } } \right) { t }^{ 2 }$$

$$\left( \cfrac { { V }^{ } }{ { T }^{ } } \right) { t }^{ 2 }$$

Solution

Question 7

$$8\times 10^{-3}s$$

$$8\times 10^{-4}s$$

$$80\times 10^{-3}s$$

$$800\times 10^{-3}s$$

Solution

Question 8

$$\dfrac {t_{1} + t_{2}}{2}$$

$$\dfrac {t_{1} - t_{2}}{2}$$

$$\sqrt {t_{1}t_{2}}$$

$$t_{1} + t_{2}$$

Solution

Question 9

$$2\ s$$

$$2.5\ s$$

$$4\ s$$

$$4.5\ s$$

$$1\ s$$

Solution

Question 10

27 km/hr

144 km/hr

72 km/hr

83 km/hr

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.