Motion in a Straight Line — Page 10 | NEET Physics

Q91

The position of a particle related to time is given by

x=\left(5 t^{2}-4 t+5\right) \mathrm{m}

. The magnitude of velocity of the particle at

t=2 s

will be :

A

14 \mathrm{~ms}^{-1}

B

16 \mathrm{~ms}^{-1}

C

10 \mathrm{~ms}^{-1}

D

06 \mathrm{~ms}^{-1}

Correct Answer

Option B

Solution

The position of a particle as a function of time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$ .

To find the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ , we first need to find the velocity of the particle as a function of time.

The velocity $v$ is the time derivative of the position $x$ :

v = \frac{dx}{dt}

Taking the derivative of $x$ with respect to $t$ , we get:

v = \frac{dx}{dt} = 10t - 4\,\mathrm{m/s}

Now we can find the velocity of the particle at $t=2\,\mathrm{s}$ by plugging in $t=2$ :

v(2\,\mathrm{s}) = 10(2) - 4\,\mathrm{m/s} = 16\,\mathrm{m/s}

Therefore, the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ is:

\boxed{|v(2\,\mathrm{s})| = 16\,\mathrm{m/s}}

Q92

A particle moves from the point

\left( {2.0\widehat i + 4.0\widehat j} \right)

m, at t = 0, with an initial velocity

\left( {5.0\widehat i + 4.0\widehat j} \right)

ms

-

1. It is acted upon by a constant force which produces a constant acceleration

\left( {4.0\widehat i + 4.0\widehat j} \right)

ms

-

2. What is the distance of the particle from the origin at time 2 s?

A 15 m

B

20\sqrt 2

m

C

10\sqrt 2

m

D 5 m

Correct Answer

Option B

Solution

\overrightarrow S = \left( {5\widehat i + 4} \right)2 + {1 \over 2}\left( {4\widehat i + 4\widehat j} \right)4

= 10\widehat i + 8\widehat j + 8\widehat i + 8\widehat j

\overrightarrow {{r_f}} - \overrightarrow {{r_i}} = 18\widehat i + 16\widehat j

\overrightarrow {{r_f}} = 20\widehat i + 20\widehat j

\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2

Q93

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to : [g is the acceleration due to gravity]

A t = 3.4

\sqrt {\left( {{h \over g}} \right)}

B t = 1.8

\sqrt {\left( {{h \over g}} \right)}

C t =

\sqrt {{{2h} \over {3g}}}

D t =

{2 \over 3}\sqrt {\left( {{h \over g}} \right)}

Correct Answer

Option A

Solution

Velocity of helicopter at height h,

V_B^2 = {0^2} + 2gh

{V_B} = \sqrt {2gh}

- h = ({V_B})t - {1 \over 2}g{t^2}

$\Rightarrow$

- h = \sqrt {2ght} - {1 \over 2}g{t^2}

$\Rightarrow$

g{t^2} - 2\sqrt {2ght} - 2h = 0

$\Rightarrow$

t = {{\sqrt {2ght} \pm \sqrt {8gh + 8gh} } \over {2g}} = {{2\sqrt {2gh} \pm \sqrt {16gh} } \over {2g}} = {{\sqrt {2gh} + 2\sqrt {gh} } \over g}

$\Rightarrow$

t = \sqrt {{{2h} \over g}} + 2\sqrt {{h \over g}} = \sqrt {{h \over g}} \left( {\sqrt 2 + 2} \right) = 3.4\sqrt {{h \over g}}

Q94

A vehicle travels

4 \mathrm{~km}

with speed of

3 \mathrm{~km} / \mathrm{h}

and another

4 \mathrm{~km}

with speed of

5 \mathrm{~km} / \mathrm{h}

, then its average speed is

A

3.75 \mathrm{~km} / \mathrm{h}

B

4.25 \mathrm{~km} / \mathrm{h}

C

3.50 \mathrm{~km} / \mathrm{h}

D

4.00 \mathrm{~km} / \mathrm{h}

Correct Answer

Option A

Solution

Average speed

= {{\mathrm{Total\,dis\tan ce}} \over {\mathrm{Total\,time}}}

= {8 \over {{4 \over 3} + {4 \over 5}}}

= 3.75

km/h

Q95

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

A 2gH = n2u2

B gH = (n - 2)2u2

C 2gH = nu2(n - 2)

D gH = (n - 2)u2

Correct Answer

Option C

Solution

Time taken to reach highest point is

t = {u \over g}

Time taken by the particle to reach the ground =

nt = {\nu \over g}

Speed on reaching ground

v = \sqrt {{u^2} + 2gH}

Now,

v = u + at

\Rightarrow \sqrt {{u^2} + 2gH} = - u + gt

\Rightarrow t = {{u + \sqrt {{u^2} + 2gH} } \over g} = {{\nu} \over g}

(from question)

\Rightarrow 2gH = n\left( {n - 2} \right){u^2}

Q96

A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to

{{81} \over {100}}

of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms

-

2)

A 2.50 ms

-

1

B 3.0 ms

-

1

C 2.0 ms

-

1

D 3.50 ms

-

1

Correct Answer

Option A

Solution

Total distance d = h + 2e2h + 2e4h + 2e6h + 2e8h + ...... d = h + 2e2h (1 + e2 + e4 + e6 + .......)

d = {{(1 - {e^2})h + 2{e^2}h} \over {1 - {e^2}}} = {{h(1 + {e^2})} \over {1 - {e^2}}}

Total time = T + 2eT + 2e2T + 2e3T + ....... Total time = T + 2eT (1 + e + e2 + e3 + .......) =

T + 2e.T\left( {{1 \over {1 - e}}} \right)

Total time

= {{T(1 + e)} \over {1 - e}}

Average speed of the ball

{V_{avg}} = {{h{{(1 + {e^2})} \over {(1 - {e^2})}}} \over {T\left( {{{1 + e} \over {1 - e}}} \right)}}

= {5 \over 1}\left( {{{1 + {e^2}} \over {(1 + e)(1 - e)}}{{(1 - e)} \over {(1 + e)}}} \right)

{V_{avg}} = {{5(1 + {e^2})} \over {{{(1 + e)}^2}}}

$\because$

{h^1} = {e^2}h

{{81} \over {100}} = {e^2}

e = {9 \over {10}} = 0.9

{V_{avg}} = {{5\left( {1 + {{81} \over {100}}} \right)} \over {{{(1 + 0.9)}^2}}}

= 2.50 m/sec.

Q97

A car starting from rest accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate

{f \over 2}

to come to rest. If the total distance traversed is 15 S, then

A

S = {1 \over 6}f{t^2}

B

S = ft

C

S = {1 \over 4}f{t^2}

D

S = {1 \over 72}f{t^2}

Correct Answer

Option D

Solution

Initially car starts from rest so u = 0. Now distance from

A

to

B

,

\,\,\,\,\,\,\,\,\,\,

S = {1 \over 2}ft_1^2

\Rightarrow ft_1^2 = 2S

Distance from

B

to

C

= \left( {f{t_1}} \right)t

In B to C velocity is constant and v =

{f{t_1}}

Distance from

C

to

D

\,\,\,\,\,\,\,\,\,\,

= {{{u^2}} \over {2a}} = {{{{\left( {f{t_1}} \right)}^2}} \over {2\left( {f/2} \right)}} = ft_1^2 = 2S

\Rightarrow S + f\,{t_1}t + 2S = 15S

\Rightarrow f\,{t_1}t = 12S

........(1) But

\,\,\,\,\,\,\,\,\,

{1 \over 2}f\,t_1^2 = S

.........(2) On dividing the above two equations, we get

{t_1} = {t \over 6}

\Rightarrow S = {1 \over 2}f{\left( {{t \over 6}} \right)^2} = {{f\,{t^2}} \over {72}}

Q98

A car travels a distance of '

x

' with speed

v_1

and then same distance '

x

' with speed

v_2

in the same direction. The average speed of the car is :

A

{{{v_1}{v_2}} \over {2({v_1} + {v_2})}}

B

{{2{v_1}{v_2}} \over {{v_1} + {v_2}}}

C

{{2x} \over {{v_1} + {v_2}}}

D

{{{v_1} + {v_2}} \over 2}

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Average velocity }=\frac{\text{ Total displacement }}{\text{ Total time }} \\\\ & =\frac{\mathrm{x}+\mathrm{x}}{\frac{\mathrm{x}}{\mathrm{v}_1}+\frac{\mathrm{x}}{\mathrm{v}_2}}=\frac{2 \mathrm{v}_1 \mathrm{v}_2}{\mathrm{v}_1+\mathrm{v}_2} \end{aligned}