Variables of Motion questions | Variables of Motion jee questions

Q 1 :
A bullet is fired into a fixed target looses one third of its velocity after travelling 4cm. It penetrates further $D \times 10^{- 3}$ m before coming to rest. The value of D is [2024]

32
5
3
4

1

2

3

4

(A) $v^{2} - u^{2} = 2 a s$

${(\frac{2 u}{3})}^{2} = u^{2} + 2 (- a) (4 \times 10^{- 2})$

$\frac{4 u^{2}}{9} = u^{2} - 2 a (4 \times 10^{- 2})$

$- \frac{5 u^{2}}{9} = - 2 a (4 \times 10^{- 2})$ ...(1)

$0 = {(\frac{2 u}{3})}^{2} + 2 (- a) (x)$

$- \frac{4 u^{2}}{9} = - 2 a x$ ...(2)

$(1) / (2), \frac{5}{4} = \frac{4 \times 10^{- 2}}{x}$

$x = \frac{16}{5} \times 10^{- 2}$

$x = 3.2 \times 10^{- 2} m$

Q 2 :
A particle is moving in a straight line. The variation of position $' x'$ as a function of time $' t'$ is given as $x = (t^{3} - 6 t^{2} + 20 t + 15)$ m. The velocity of the body when its acceleration becomes zero is [2024]

6 m/s
10 m/s
8 m/s
4 m/s

1

2

3

4

(C) $x = t^{3} - 6 t^{2} + 20 t + 15$

$\frac{d x}{d t} = v = 3 t^{2} - 12 t + 20$

$\frac{d v}{d t} = a = 6 t - 12$

When $a = 0$

$t = 2 s e c$

At $t = 2 s e c$

$v = 3 {(2)}^{2} - 12 (2) + 20 = 8 m / s$

Q 3 :
The relation between time $' t'$ and distance $' x'$ is $t = α x^{2} + β x,$ where $α$ and $β$ are constants. The relation between acceleration (a) and velocity (v) is [2024]

$a = - 2 α v^{3}$
$a = - 2 α v^{4}$
$a = - 2 α v^{2}$
$a = - 2 α v^{5}$

1

2

3

4

(A) $t = α x^{2} + β x$ (differentiating wrt time)

$\frac{d t}{d x} = 2 α x + β$

$\frac{1}{v} = 2 α x + β$

(differentiating wrt time) $- \frac{1}{v^{2}} \frac{d v}{d t} = 2 α \frac{d x}{d t}$

$\frac{d v}{d t} = - 2 α v^{3}$

Q 4 :
A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^{2} = 1 + t^{2}$ . Its acceleration at any time $t$ is $x^{- n}$ where $n =$ _____. [2024]

0%

(3) $x^{2} = 1 + t^{2} \Rightarrow x = \sqrt{1 + t^{2}}$

$v = \frac{d x}{d t} = \frac{1}{2 \sqrt{1 + t^{2}}} \cdot 2 t \Rightarrow v = \frac{t}{\sqrt{1 + t^{2}}}$

$a = \frac{d v}{d t} = \frac{\sqrt{1 + t^{2}} (1) - t \cdot \frac{1}{2} {(t^{2} + 1)}^{- 1 / 2} \cdot 2 t}{(1 + t^{2})}$

$a = \frac{\sqrt{1 + t^{2}} - \frac{t^{2}}{\sqrt{1 + t^{2}}}}{(1 + t^{2})} = \frac{t^{2} + 1 - t^{2}}{{(t^{2} + 1)}^{3 / 2}}$

$\Rightarrow a = \frac{1}{{(1 + t^{2})}^{3 / 2}} = \frac{1}{x^{3}} = x^{- 3} \Rightarrow n = 3$

Q 5 :
A particle is moving in one dimension (along x-axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as $x = - 3 t^{3} + 18 t^{2} + 16 t,$ where x is in m and t in s. The velocity of the particle when its acceleration becomes zero is _____ m/s. [2024]

0%

(52) $x = 3 t^{3} + 18 t^{2} + 16 t$

$v = - 9 t^{2} + 36 t + 16$

$a = - 18 t + 36$

$a = 0$ at $t = 2$ $s$

$v = - 9 {(2)}^{2} + 36 \times 2 + 16$

$v = 52 m / s$

Q 6 :
A particle initially at rest starts moving from reference point. $x = 0$ along x-axis, with velocity $v$ that varies as $v = 4 \sqrt{x}$ m/s. The acceleration of the particle is _____ $m s^{- 2}$ . [2024]

0%

(8) Given, $v = 4 \sqrt{x}$

$‐ ‐ ‐ □ ‐ ‐ ‐ \to ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ x - a x i s$

$x = 0, t = 0$

$a = \frac{v d v}{d x} = (4 \sqrt{x}) \frac{d (4 \sqrt{x})}{d x} = 16 \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}$

$16 \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} = 8 {m / s}^{2}$

Topic Question Set

Q 1 :
A bullet is fired into a fixed target looses one third of its velocity after travelling 4cm. It penetrates further $D \times 10^{- 3}$ m before coming to rest. The value of D is [2024]

Q 2 :
A particle is moving in a straight line. The variation of position $' x'$ as a function of time $' t'$ is given as $x = (t^{3} - 6 t^{2} + 20 t + 15)$ m. The velocity of the body when its acceleration becomes zero is [2024]

Q 3 :
The relation between time $' t'$ and distance $' x'$ is $t = α x^{2} + β x,$ where $α$ and $β$ are constants. The relation between acceleration (a) and velocity (v) is [2024]

Q 4 :
A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^{2} = 1 + t^{2}$ . Its acceleration at any time $t$ is $x^{- n}$ where $n =$ _____. [2024]

0%

0%

Q 6 :
A particle initially at rest starts moving from reference point. $x = 0$ along x-axis, with velocity $v$ that varies as $v = 4 \sqrt{x}$ m/s. The acceleration of the particle is _____ $m s^{- 2}$ . [2024]

0%

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : A bullet is fired into a fixed target looses one third of its velocity after travelling 4cm. It penetrates further D×10-3 m before coming to rest. The value of D is [2024]

Q 2 : A particle is moving in a straight line. The variation of position 'x' as a function of time 't' is given as x=(t3-6t2+20t+15)m. The velocity of the body when its acceleration becomes zero is [2024]

Q 3 : The relation between time 't' and distance 'x' is t=αx2+βx, where α and β are constants. The relation between acceleration (a) and velocity (v) is [2024]

Q 4 : A particle moves in a straight line so that its displacement x at any time t is given by x2=1+t2. Its acceleration at any time t is x-n where n=_____. [2024]

0%

0%

Q 6 : A particle initially at rest starts moving from reference point. x=0 along x-axis, with velocity v that varies as v=4x m/s. The acceleration of the particle is _____ ms-2. [2024]

0%

Want Us to Email you About Special Offers & Updates?

Q 1 :
A bullet is fired into a fixed target looses one third of its velocity after travelling 4cm. It penetrates further $D \times 10^{- 3}$ m before coming to rest. The value of D is [2024]

Q 2 :
A particle is moving in a straight line. The variation of position $' x'$ as a function of time $' t'$ is given as $x = (t^{3} - 6 t^{2} + 20 t + 15)$ m. The velocity of the body when its acceleration becomes zero is [2024]

Q 3 :
The relation between time $' t'$ and distance $' x'$ is $t = α x^{2} + β x,$ where $α$ and $β$ are constants. The relation between acceleration (a) and velocity (v) is [2024]

Q 4 :
A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^{2} = 1 + t^{2}$ . Its acceleration at any time $t$ is $x^{- n}$ where $n =$ _____. [2024]

Q 6 :
A particle initially at rest starts moving from reference point. $x = 0$ along x-axis, with velocity $v$ that varies as $v = 4 \sqrt{x}$ m/s. The acceleration of the particle is _____ $m s^{- 2}$ . [2024]