Variables of Motion questions | Variables of Motion jee questions

Q 11 :

A person travelling on a straight line moves with a uniform velocity $v_{1}$ for a distance x and with a uniform velocity $v_{2}$ for the next $\frac{3}{2} x$ distance. The average velocity in this motion is $\frac{50}{7}$ m/s. If $v_{1}$ is 5 m/s then $v_{2}$ = __________ m/s. [2025]

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(10)

$v_{avg} = \frac{x_{1} + x_{2}}{t_{1} + t_{2}}$

$\Rightarrow \frac{50}{7} = \frac{x + \frac{3 x}{2}}{\frac{x}{5} + \frac{3 x}{2 v_{2}}}$

$\Rightarrow \frac{50}{7} = \frac{5 / 2}{\frac{1}{5} + \frac{3}{2 v_{2}}}$

$\Rightarrow v_{2} = 10 m / s$

Q 12 :

The distance travelled by a particle is related to time t as $x = 4 t^{2}$ . The velocity of the particle at t = 5 s is [2023]

8 ${ms}^{- 1}$
20 ${ms}^{- 1}$
25 ${ms}^{- 1}$
40 ${ms}^{- 1}$

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(4)

$x = 4 t^{2}$

$v = \frac{d x}{d t} = 8 t$

$At t = 5 s,$

$v = 8 \times 5 = 40 m/s$

Q 13 :

A vehicle travels 4 km with speed of 3 km/h and another 4 km with speed of 5 km/h, then its average speed is [2023]

4.25 km/h
3.50 km/h
4.00 km/h
3.75 km/h

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4

(4)

$\frac{2}{V_{av}} = \frac{1}{3} + \frac{1}{5} = \frac{8}{15}$

$\Rightarrow V_{av} = \frac{15}{4} = 3.75 km/h$

Q 14 :

An object moves with speeds $v_{1}$ , $v_{2}$ and $v_{3}$ along the line segments AB, BC and CD respectively as shown in the figure. Where AB = BC and AD = 3AB, then the average speed of the object will be: [2023]

$\frac{(v_{1} + v_{2} + v_{3})}{3}$
$\frac{v_{1} v_{2} v_{3}}{3 (v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1})}$
$\frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}$
$\frac{(v_{1} + v_{2} + v_{3})}{3 v_{1} v_{2} v_{3}}$

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(3)

$A B = x and B C = x$

$2 x + C D = 3 x$

$\Rightarrow C D = x$

$< v > = \frac{3 x}{\frac{x}{v_{1}} + \frac{x}{v_{2}} + \frac{x}{v_{3}}} = \frac{3 v_{1} v_{2} v_{3}}{v_{2} v_{3} + v_{1} v_{3} + v_{1} v_{2}}$

Q 15 :

A car travels a distance of $' x'$ with speed $v_{1}$ and then the same distance $' x'$ with speed $v_{2}$ in the same direction. The average speed of the car is: [2023]

$\frac{v_{1} v_{2}}{2 (v_{1} + v_{2})}$
$\frac{2 x}{v_{1} + v_{2}}$
$\frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$
$\frac{v_{1} + v_{2}}{2}$

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(3)

$V_{avg} = \frac{S + S}{\frac{S}{V_{1}} + \frac{S}{V_{2}}} = \frac{2 V_{1} V_{2}}{V_{1} + V_{2}}$

Q 16 :

A person travels $x$ distance with velocity $v_{1}$ and then $x$ distance with velocity $v_{2}$ in the same direction. The average velocity of the person is $v$ , then the relation between $v$ , $v_{1}$ and $v_{2}$ will be: [2023]

$v = v_{1} + v_{2}$
$v = \frac{v_{1} + v_{2}}{2}$
$\frac{2}{v} = \frac{1}{v_{1}} + \frac{1}{v_{2}}$
$\frac{1}{v} = \frac{1}{v_{1}} + \frac{1}{v_{2}}$

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(3)

$Average velocity = \frac{x + x}{\frac{x}{v_{1}} + \frac{x}{v_{2}}} = v$

$\frac{1}{v_{1}} + \frac{1}{v_{2}} = \frac{2}{v}$

Q 17 :

The distance travelled by an object in time $t$ is given by $s = (2.5) t^{2} .$ The instantaneous speed of the object at $t$ = 5 s will be: [2023]

12.5 ${ms}^{- 1}$
62.5 ${ms}^{- 1}$
5 ${ms}^{- 1}$
25 ${ms}^{- 1}$

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(4)

$Distance (s) = (2.5) t^{2}$

$Speed (v) = \frac{d s}{d t} = \frac{d}{d t} {(2.5) t^{2}}$

$v = 5 t$

$At t = 5, v = 5 \times 5 = 25 m/s$

Option (4) is correct

Q 18 :

The position of a particle related to time is given by $x = (5 t^{2} - 4 t + 5) m .$ The magnitude of velocity of the particle at $t = 2 s$ will be: [2023]

${10 ms}^{- 1}$
${14 ms}^{- 1}$
${16 ms}^{- 1}$
${6 ms}^{- 1}$

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(3)

$x = 5 t^{2} - 4 t + 5$

$v = 10 t - 4$

$At t = 2 s, v = 16 m/s$

Q 19 :

A horse rider covers half the distance with 5 m/s speed. The remaining part of the distance was travelled with speed 10 m/s for half the time and with speed 15 m/s for the other half of the time. The mean speed of the rider averaged over the whole time of motion is $x / 7$ m/s. The value of $x$ is ______. [2023]

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(50)

$t_{A B} = \frac{x}{5 m/s}$

In motion BC, $x = d_{1} + d_{2}$

where $d_{1}$ and $d_{2}$ are the distances travelled with 10 m/s and 15 m/s respectively in equal time intervals $\frac{' t'}{2}$ each.

$d_{1} = \frac{10 t}{2}, d_{2} = \frac{15 t}{2}$

$d_{1} + d_{2} = x = \frac{t}{2} (10 + 15) = \frac{25 t}{2}$

$< v > = \frac{2 x}{\frac{x}{5} + \frac{2 x}{25}} = \frac{2 \times 25}{5 + 2} = \frac{50}{7} m/s$

Topic Question Set

Q 11 :

A person travelling on a straight line moves with a uniform velocity $v_{1}$ for a distance x and with a uniform velocity $v_{2}$ for the next $\frac{3}{2} x$ distance. The average velocity in this motion is $\frac{50}{7}$ m/s. If $v_{1}$ is 5 m/s then $v_{2}$ = __________ m/s. [2025]

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Q 12 :

The distance travelled by a particle is related to time t as $x = 4 t^{2}$ . The velocity of the particle at t = 5 s is [2023]

Q 13 :

A vehicle travels 4 km with speed of 3 km/h and another 4 km with speed of 5 km/h, then its average speed is [2023]

Q 14 :

An object moves with speeds $v_{1}$ , $v_{2}$ and $v_{3}$ along the line segments AB, BC and CD respectively as shown in the figure. Where AB = BC and AD = 3AB, then the average speed of the object will be: [2023]

Q 15 :

A car travels a distance of $' x'$ with speed $v_{1}$ and then the same distance $' x'$ with speed $v_{2}$ in the same direction. The average speed of the car is: [2023]

Q 16 :

A person travels $x$ distance with velocity $v_{1}$ and then $x$ distance with velocity $v_{2}$ in the same direction. The average velocity of the person is $v$ , then the relation between $v$ , $v_{1}$ and $v_{2}$ will be: [2023]

Q 17 :

The distance travelled by an object in time $t$ is given by $s = (2.5) t^{2} .$ The instantaneous speed of the object at $t$ = 5 s will be: [2023]

Q 18 :

The position of a particle related to time is given by $x = (5 t^{2} - 4 t + 5) m .$ The magnitude of velocity of the particle at $t = 2 s$ will be: [2023]

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