Motion in A Straight Line Test

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Motion in A Straight Line Test - 25

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Question 1
1 / -0

A truck running along a straight line increases its speed uniformly from $$30$$m/s to $$60$$m/s over a time interval $$1$$ min. The distance travelled during this time interval is

A
$$900$$ m

B
$$1800$$ m

C
$$2700$$ m

D
$$3600$$ m

Solution

Acceleration of the truck $$= \dfrac{v_2 - v_1}{t_2-t_1} $$

a $$= \dfrac{60 - 30}{60} $$
$$= 0.5\ m/s^2 $$

Distance travelled during this time interval is
s $$= ut + \dfrac{1}{2}at^2 $$
$$ = 30 \times 60 + \dfrac{0.5 \times 60^2}{2}$$
$$= 2700\ m $$
Question 2
1 / -0

If a body is thrown vertically upward and rises to a height of 10 m, then time taken by the body to reach the highest point is

A
1.043 s

B
1.43 s

C
1.024 s

D
none of these

Solution

Using the equation of motions
$$v^2 = u^2 - 2gh $$
At maximum height , v = 0
$$u^2 = 2gH $$
$$u^2 = 2 \times 9.8 \times 10 $$
$$u^2 = 196 $$
$$u = 14 m/s $$

$$v = u - gt $$
$$t = \frac{u}{g} $$
$$t = \frac{14}{9.8} $$
$$ = 1.43 s$$
Question 3
1 / -0

A particle is thrown upwards from ground. It experiences a constant air resistance force which can produce a retardation of $$2 m/s^{2}.$$ The ratio of time of ascent to the time of descent is $$:\left[ g = 10\ m/s^{2} \right] $$

A
$$1 : 1$$

B
$$\sqrt{\dfrac{2}{3}}$$

C
$${\dfrac{2}{3}}$$

D
$$\sqrt{\dfrac{3}{2}}$$

Solution

Let a be the retardation produced by resistive force, $$t_{a}\ and\ t_{d}$$ be the time ascent and descent respectively.
Let the particle rise upto a height $$h$$ and projection velocity be $$u$$.

For upward motion,
$$0 = u - (a+g)t_a ..........(i)$$
$$0 = u^2 -2(a+g)h..........(ii)$$

For downward motion,
$$v_2 = 0 + (g-a)t_d.........(iii)$$
$$v_2^2 = 0 + 2(g-a)h.......(iv)$$

From $$(ii)\ \&\ (iv),$$
$$v_2 = u_2 \sqrt{\cfrac{ g-a}{g+a}}...........(v)$$

From $$(i)\ \&\ (iii),$$
$$\cfrac{t_a}{t_d}= \cfrac{u(g-a)}{v_2(g+a)}............(vi)$$

From $$(v)\ \&\ (vi),$$
$$\cfrac{t_a}{t_d} = \sqrt{\cfrac{g-a}{g+a}}$$
$$=\sqrt { \cfrac{2}{3}}$$
Question 4
1 / -0

A particle moves in a straight line with a constant acceleration. It changes its velocity from $$ 10 ms^{-1} \ to \ 20 ms^{-1}$$ while passing through a distance $$135m$$ in $$t$$ second. The value of t is

A
$$12$$

B
$$9$$

C
$$10$$

D
$$1.8$$

Solution

$$u=10m/s,v=20m/s,x=130m$$
$$a=\cfrac{v-u}{t}=\cfrac{20-10}{t}=\cfrac{10}{t}$$
$$s=ut+\cfrac{1}{2}at^2\\\implies 135 =10t+\cfrac{1}{2}\cfrac{10}{t}t^2\\\implies t=9s$$
Question 5
1 / -0

An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s.
What will be the velocity( in $$m/s$$) after 7 s from the start.

A
0

B
10

C
65

D
70

Solution

From $$s=ut+\dfrac{1}{2}at^2$$
In first 2 s,
$$s=20 m$$
$$20 =\frac{1}{2}a(2)^2=2 a$$
$$a = 10\ m/s^2$$
Velocity at the end of 2 s is
$$v = u+at = 0 + 10 \times 2 = 20 m/s$$
In next 4s, $$s=ut +\dfrac{1}{2} a't^2$$
$$160 = 20 \times 4+\frac{1}{2}a'(4)^2$$
$$80 = 8 a'$$
$$a' = 10\ m/s^2$$
It shows that acceleration is uniform. From $$v=u + at$$
$$ v=0+10 \times 7 = 70\ m/s$$
Answer: (D) 70
Question 6
1 / -0

A car starts from rest and moves along the x-axis with a constant acceleration $$5\ ms^{-2}$$ for $$8\ s$$. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?

A
160 m

B
240 m

C
320 m

D
400 m

Solution

The distance travelled in first eight seconds $$=x_{1}=0+\frac{1}{2}*5*8^{2}=160\ m$$
At this point the velocity $$ v=u+at=0+ 5\times 8 =40\ ms^{-1}$$
Therefore, the distance covered in last four seconds, $$x_{2}= 40 \times 4=160 m$$
Thus, the total distance $$x=x_{1}+x_{2}= 160+160 =320 m$$
Question 7
1 / -0

A particle is projected vertically upwards with a speed $$u$$.

A
When it rises to half its maximum height its velocity is $$\dfrac{u}{2}$$.

B
Time taken to rise to half its maximum height is half the time taken to reach maximum height.

C
Time taken to rise to three fourth its maximum height is half the time taken to reach its maximum height.

D
The acceleration of the particle when it reaches its maximum height is zero.

Solution

If $$ h$$ is the maximum height then $${ h= }\dfrac { { u }^{ 2 } }{ 2g } $$
Time taken to reach height $$ h$$, $$ t=\dfrac { u }{ g } $$
Let $$v$$ be the speed at $$ \dfrac{3h}{4}$$ then $$ { v }^{ 2 }={ u }^{ 2 }-2g\left( \dfrac { 3h }{ 4 } \right) \Rightarrow v=\dfrac { u }{ 2 } $$
let $$ t'$$ be the time taken to reach $$ \dfrac{3h}{4}$$ then $$ \dfrac { u }{ 2 } =u-gt'\Rightarrow t'=\dfrac { u }{ 2g } =\dfrac { t }{ 2 } $$
Question 8
1 / -0

A stone is dropped from a certain height, which can reach the ground in 5 s. After 3 s of its fall, it is stopped and again allowed to fall. Then, the time taken by the stone to reach the ground for the remaining distance is :

A
3 s

B
4 s

C
2 s

D
None of these

Solution

$$T_d = 5s, H= \displaystyle \dfrac{1}{2} \times g \times 25 = 12.5 g$$
$$t_1 = 3s ; h = \displaystyle \dfrac{1}{2} \times g \times 9 =4.5 g$$
$$(H-h)=8g = \displaystyle \dfrac{1}{2} \times g \times t^2 \Rightarrow t = \sqrt{16} = 4 s$$
Question 9
1 / -0

A body falls from a height of 200 m. If gravitational attraction ceases after 2 s, further time taken by it to reach the ground is $$(g = 10 m s^{-2})$$.

A
5 s

B
9 s

C
13 s

D
17 s

Solution

$$t= 2s$$

$$h_1 = \displaystyle \dfrac{1}{2} gt^2$$

$$h_1 \displaystyle = \dfrac{1}{2} \times 10 \times 4 = 20 m$$

After falling through 20 m, g becomes zero remaining distance to be travelled
(d$$ = 200 - 20 = 180 m$$)
and
$$v = gt = 10 \times 2 = 20 m s^{-1}$$

$$\therefore t \displaystyle = \dfrac{d}{v} = \dfrac{180}{20} = 9s$$
Question 10
1 / -0

A body thrown vertically up with a velocity 'u' reaches the maximum height 'h' after 'T' second. The correct statement among the following is :

A
At a height $$\displaystyle \dfrac{h}{2}$$ from the ground its velocity is $$\displaystyle \dfrac{u}{2}$$

B
At a time T its velocity is 'u'

C
At a time '2T' its velocity is -u

D
At a time 2T its velocity is -6u

Solution

Time taken by the body to reach the maximum height is given as $$T$$ and the velocity of the body at the time of throw is $$u$$ (upwards). After further time $$T$$, the body reaches to the ground having same velocity $$u$$ in the downward direction as that of initial velocity at the time of throw. Thus, at the instant $$2T$$, the velocity of the body is $$-u$$ (where minus sign represents the velocity to be in downward direction).

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 25

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Motion in A Straight Line Test - 25

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SHARING IS CARING

Question 1

$$900$$ m

$$1800$$ m

$$2700$$ m

$$3600$$ m

Solution

Question 2

1.043 s

1.43 s

1.024 s

none of these

Solution

Question 3

$$1 : 1$$

$$\sqrt{\dfrac{2}{3}}$$

$${\dfrac{2}{3}}$$

$$\sqrt{\dfrac{3}{2}}$$

Solution

Question 4

$$12$$

$$9$$

$$10$$

$$1.8$$

Solution

Question 5

0

10

65

70

Solution

Question 6

160 m

240 m

320 m

400 m

Solution

Question 7

When it rises to half its maximum height its velocity is $$\dfrac{u}{2}$$.

Time taken to rise to half its maximum height is half the time taken to reach maximum height.

Time taken to rise to three fourth its maximum height is half the time taken to reach its maximum height.

The acceleration of the particle when it reaches its maximum height is zero.

Solution

Question 8

3 s

4 s

2 s

None of these

Solution

Question 9

5 s

9 s

13 s

17 s

Solution

Question 10

At a height $$\displaystyle \dfrac{h}{2}$$ from the ground its velocity is $$\displaystyle \dfrac{u}{2}$$

At a time T its velocity is 'u'

At a time '2T' its velocity is -u

At a time 2T its velocity is -6u

Solution

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