Motion in a Straight Line — Page 4 | JEE Physics

Q31

A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is

A 12 m

B 18 m

C 24 m

D 6 m

Correct Answer

Option C

Solution

For case 1 : u = 50 km/hr =

{{50 \times 1000} \over {3600}}

m/s =

{{125} \over 9}

m/s, v = 0, s = 6 m,

a

= ? $\therefore$ 02 = u2 + 2

a

s $\Rightarrow$

a = - {{{u^2}} \over {2s}}

$\Rightarrow$

a = - {{{{\left( {{{125} \over 9}} \right)}^2}} \over {2 \times 6}}

= $-$ 16 m/s2 For case 2 : u = 100 km/hr =

{{100 \times 1000} \over {3600}}

m/s =

{{250} \over 9}

m/s, v = 0,

a

= $-$ 16, s = ? $\therefore$ 02 = u2 + 2

a

s $\Rightarrow$

s = - {{{u^2}} \over {2a}}

$\Rightarrow$

s = - {{{{\left( {{{250} \over 9}} \right)}^2}} \over {2 \times -16}}

= 24 m

Q32

A ball is released from a height h. If

t_{1}

and

t_{2}

be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between

t_{1}

and

t_{2}

.

A

t_{1}=(\sqrt{2}) t_{2}

B

t_{1}=(\sqrt{2}-1) t_{2}

C

t_{2}=(\sqrt{2}+1) t_{1}

D

t_{2}=(\sqrt{2}-1) t_{1}

Correct Answer

Option D

Solution

For first

{h \over 2}

distance,

{h \over 2} = {1 \over 2}~gt_1^2

\Rightarrow h = gt_1^2

For total distance h,

h = {1 \over 2}g{({t_1} + {t_2})^2}

\Rightarrow gt_1^2 = {1 \over 2}g{({t_1} + {t_2})^2}

\Rightarrow 2t_1^2 = {({t_1} + {t_2})^2}

\Rightarrow \sqrt 2 {t_1} = {t_1} + {t_2}

\Rightarrow (\sqrt 2 - 1){t_1} = {t_2}

Q33

A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10m in t s, the distance travelled by the toy in the next t s will be :

A 10 m

B 20 m

C 30 m

D 40 m

Correct Answer

Option C

Solution

A small toy begins to move from a standstill with a constant acceleration.

It covers a distance of 10 meters in t seconds.

We need to determine the distance the toy will travel in the subsequent t seconds.

First, we know that the initial distance traveled is given by the formula for constant acceleration starting from rest:

\frac{1}{2} a t^2 = 10 \text{ m}

Next, we calculate the total distance traveled after 2t seconds:

\frac{1}{2} a (2t)^2 = \frac{1}{2} a \cdot 4t^2 = 2 a t^2

Since we know from the initial condition that $\dfrac{1}{2} a t^2 = 10 \text{ m}$ , multiplying it by 4 gives

2 a t^2 = 40 \text{ m}

Thus, the additional distance traveled in the next t seconds is:

\text{Total distance after 2t seconds} - \text{Distance already traveled in t seconds}

= 40 \text{ m} - 10 \text{ m}

= 30 \text{ m}

Q34

The distance travelled by a particle is related to time t as

x=4\mathrm{t}^2

. The velocity of the particle at t=5s is :-

A

\mathrm{25~ms^{-1}}

B

\mathrm{20~ms^{-1}}

C

\mathrm{8~ms^{-1}}

D

\mathrm{40~ms^{-1}}

Correct Answer

Option D

Solution

\begin{aligned} & x=4 t^2 \\\\ & v=\frac{d x}{d t}=8 t \end{aligned}

At $\mathrm{t}=5 ~\mathrm{sec}$

\mathrm{v}=8 \times 5=40 \mathrm{~m} / \mathrm{s}

Q35

A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2 . The car will catch up with the bus after a time of :

A

\sqrt {110} \,s

B

\sqrt {120} \,s

C

10\,\,\sqrt 2 \,s

D 15 s

Correct Answer

Option C

Solution

Acceleration of Car, aC = 4 m/s2 Acceleration of bus, aB = 2 m/s2 Initial distance between them, S = 200 m Acceleration of Car with respect to bus, aCB = aC $-$ aB = 4 $-$ 2 = 2 m/s2 As initially both are at rest so, uCB = 0

\therefore\,\,\,

S = UCB $\times$ t +

{1 \over 2}

aCB $\times$ t2 $\Rightarrow$

\,\,\,

200 = 0 +

{1 \over 2}

$\times$ 2 $\times$ t2 $\Rightarrow$

\,\,\,

t2 = 200 $\Rightarrow$

\,\,\,

t = 10

\sqrt 2

sec.

Q36

A train starting from rest first accelerates uniformly up to a speed of

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

for time

t

, then it moves with a constant speed for time

3 t

. The average speed of the train for this duration of journey will be (in

\mathrm{km} / \mathrm{h}

) :

A 70

B 40

C 30

D 80

Correct Answer

Option A

Solution

To find the average speed of the train for the duration of the journey, we need to know the total distance covered by the train and the total time taken.

The train accelerates uniformly to a speed of

80 \, \mathrm{km/h}

over time

t

, and then moves at this constant speed for

3t

. The average speed can be calculated using the formula:

\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}

Step 1: Calculate the distance covered during acceleration The distance covered while the train is accelerating can be found using the formula for the distance travelled under uniform acceleration:

d_1 = \frac{1}{2} at^2

Where: $d_1$ is the distance covered during acceleration. $a$ is the acceleration (we don't have a direct value for this, but we can work with the given information). $t$ is the time.

However, to proceed with the calculation without the acceleration ( $a$ ), we recognize that the formula directly correlates to distance but requires knowledge of acceleration.

Instead, let's think in terms of the final speed and time, given that the train reaches

80 \, \mathrm{km/h}

(or

\frac{80}{3.6} = 22.22 \, \mathrm{m/s}

) in time

t

. Using the relationship between velocity, time, and distance, since the acceleration is uniform, we can use:

d_1 = v \times t_1 - \frac{1}{2} a t^2

Given that the initial speed $u = 0$ and final speed $v = 80 \, \mathrm{km/h}$ , converting the speed to meters per second (since our time is likely in seconds) gives us $22.22 \, \mathrm{m/s}$ .

But without directly calculating acceleration, we simplify using average speed for the acceleration phase because it starts from rest and reaches $v$ .

The average speed during acceleration, \(v_{avg} = \frac{u + v}{2} = \frac{0 + 80}{2} = 40 \, \mathrm{km/h}

. Thus, the distance d_1 = v_{avg} \times t = 40 \, \mathrm{km/h} \times t. Step 2: Calculate the distance covered at constant speed The distance covered at a constant speed is easier to calculate:

d_2 = v \times t_2 = 80 \, \mathrm{km/h} \times 3t

Step 3: Calculate the total distance and the total time The total distance (D) covered is the \sum of d_1 and d_2:

D = d_1 + d_2 = 40t + 240t = 280t \, \mathrm{km}

The total time (T) taken is t + 3t = 4t. Step 4: Calculate the average speed Substitute the values of D and T in the formula of average speed:

\text{Average speed} = \frac{280t}{4t}$ $This simplifies to$ 70 \, \mathrm{km/h} $. So, the average speed of the train for this duration of the journey is$ 70 \, \mathrm{km/h}$, which matches with Option A.

Q37

A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in

{T \over 3}

seconds?

A

{{8h} \over 9}

meters from the ground

B

{{7h} \over 9}

meters from the ground

C

{h \over 9}

meters from the ground

D

{{7h} \over {18}}

meters from the ground

Correct Answer

Option A

Solution

We know that equation of motion,

s = ut + {1 \over 2}g{t^2},\,\,

Initial speed of ball is zero and it take T second to reach the ground. $\therefore$

h = {1 \over 2}g{T^2}

After

T/3

second, vertical distance moved by the ball

h' = {1 \over 2}g{\left( {{T \over 3}} \right)^2}

\Rightarrow h' = {1 \over 2} \times {{8{T^2}} \over 9}

= {h \over 9}

$\therefore$ Height from ground

= h - {h \over 9} = {{8h} \over 9}

Q38

The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

A v0 + g/2 + f

B v0 + 2g + 3f

C v0 + g/2 + f/3

D v0 + g + f

Correct Answer

Option C

Solution

Given that, v = v0 + gt + ft2 We know that,

v = {{dx} \over {dt}}

\Rightarrow dx = v\,dt

Integrating,

\int\limits_0^x {dx} = \int\limits_0^t {v\,dt}

or

\,\,\,\,\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)} dt

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {{v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}} \right]_0^t

or

\,\,\,\,\,x = {v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}

At

t = 1,\,\,\,\,\,x = {v_0} + {g \over 2} + {f \over 3}.

Q39

From a building two balls A and B are thrown such that A is thrown upwards and B downwards ( both vertically with the same speed ). If vA and vB are their respective velocities on reaching the ground, then

A

{v_B} > {v_A}

B

{v_A} = {v_B}

C

{v_A} > {v_B}

D their velocities depend on their masses.

Correct Answer

Option B

Solution

Assume the initial velocity of each particle is = u And height of building = h If final velocity of A is vA then vA2 = u2 + 2(-g)(-h) = u2 + 2gh If final velocity of B is vB then vB2 = u2 + 2gh $\therefore$ vA = vB Sign Rule : Take the direction of initial velocity positive opposite direction as negative.

Here for ball A initial velocity u is upward so upward is positive and downward is negative.

That is why gravity is = - g and height = - h And for ball B initial velocity u is downward so downward is positive and upward is negative.

That is why gravity is = + g and height = + h

Q40

A particle starts with an initial velocity of

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

along

x

-direction and accelerates uniformly at the rate of

2.0 \mathrm{~ms}^{-2}

. The time taken by the particle to reach the velocity of

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

is __________.

A 30s

B 6s

C 3s

D 25s

Correct Answer

Option D

Solution

To find the time taken by the particle to reach the velocity of

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

, we can use the formula:

v = u + at

Where:

v

is the final velocity,

u

is the initial velocity,

a

is the acceleration, and

t

is the time taken. Plugging in the given values:

60.0 \mathrm{~ms}^{-1} = 10.0 \mathrm{~ms}^{-1} + 2.0 \mathrm{~ms}^{-2} \cdot t

Solve for

t

:

50.0 \mathrm{~ms}^{-1} = 2.0 \mathrm{~ms}^{-2} \cdot t

t = \frac{50.0 \mathrm{~ms}^{-1}}{2.0 \mathrm{~ms}^{-2}} = 25 \mathrm{s}

So, the time taken by the particle to reach the velocity of

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

is 25 seconds.