Motion in a Straight Line questions | Motion in a Straight Line jee questions

Q 1 :

A body travels 102.5 m in $n^{t h}$ second and 115.0 m in ${(n + 2)}^{t h}$ second. The acceleration is [2024]

$5 m / s^{2}$
$12.5 m / s^{2}$
$6.25 m / s^{2}$
$9 m / s^{2}$

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(3)

$S_{n t h} = u + \frac{a}{2} (2 n - 1)$

$102.5 = u + \frac{a}{2} (2 n - 1)$

$115 = u + \frac{a}{2} (2 n + 4 - 1)$

Equation (ii) - (i)

$115 - 102.5 = \frac{a}{2} {2 n + 3 - 2 n + 1} = \frac{a}{2} \times 4 = 2 a$

$\Rightarrow a = \frac{12.5}{2} = 6.25 m / s^{2}$

Q 2 :

A body starts moving from rest with constant acceleration covers displacement $S_{1}$ in first $(p - 1)$ seconds and $S_{2}$ in first $p$ seconds. The displacement $S_{1} + S_{2}$ will be made in time [2024]

$(2 p + 1) s$
$\sqrt{(2 p^{2} - 2 p + 1)} s$
$(2 p - 1) s$
$(2 p^{2} - 2 p + 1) s$

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(2)

The displacement $S_{1}$ in first $(p - 1)$ sec, $S_{1} = \frac{1}{2} a {(p - 1)}^{2}$

The displacement $S_{2}$ in first $p$ sec, $S_{2} = \frac{1}{2} a {(p)}^{2}$

$S_{1} + S_{2} = \frac{1}{2} a t^{2}$

${(p - 1)}^{2} + p^{2} = t^{2} \Rightarrow t = \sqrt{2 p^{2} + 1 - 2 p}$

Q 3 :

A bus moving along a straight highway with speed of 72 km/h is brought to halt within 4 s after applying the brakes. The distance travelled by the bus during this time (assume the retardation is uniform) is _______ m. [2024]

0%

(40) $Initial velocity = u = 72 km / h = 20 m / s$

$v = u + a t$

$\Rightarrow 0 = 20 + a \times 4$

$\Rightarrow a = - 5 {m / s}^{2}$

$Now, v^{2} = u^{2} + 2 a s$

$\Rightarrow 0 = 400 - 10 x$

$\Rightarrow x = 40 m$

Q 4 :

A body moves on a frictionless plane starting from rest. If $S_{n}$ is distance moved between $t = n - 1$ and $t = n$ and $S_{n - 1}$ is distance moved between $t = n - 2$ and $t = n - 1,$ then the ratio $\frac{S_{n - 1}}{S_{n}}$ is $(1 - \frac{2}{x})$ for $n = 10 .$ The value of $x$ is __________ . [2024]

0%

(19) $S_{n} = \frac{1}{2} a {(n)}^{2} - \frac{1}{2} a {(n - 1)}^{2} = \frac{1}{2} a (2 n - 1)$

$S_{n - 1} = \frac{1}{2} a (2 \times (n - 1) - 1) = \frac{1}{2} a (2 n - 3)$

$\frac{S_{n - 1}}{S_{n}} = \frac{2 n - 3}{2 n - 1} = 1 - \frac{2}{2 n - 1} = 1 - \frac{2}{20 - 1} = 1 - \frac{2}{19}$

Q 5 :

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t + 1)$ s are 125 m and 50 m/s, respectively. The distance travelled by the particle in ${(t + 2)}^{t h} s$ is __________ m. [2024]

0%

(175) $a = \frac{Δ V}{Δ t} = \frac{50}{1} = 50 {m / s}^{2}$

$displacement in {(t + 1)}^{th} s e c o n d$

$S_{1} = 125 m$

$\Rightarrow 125 = u + \frac{50}{2} [2 (t + 1) - 1]$

$\Rightarrow u = 125 - 25 [2 t + 1]$ ... (1)

$displacement in {(t + 2)}^{th} second$

$S_{2} = u + \frac{50}{2} [2 (t + 2) - 1] = u + 25 [2 t + 3]$ ...(2)

Using equation (1) in equation (2)

$S_{2} = 125 - 25 (2 t + 1) + 25 (2 t + 3)$

$= 125 - 25 + 75 = 175 m$

Q 6 :

A particle starts with an initial velocity of 10.0 ${ms}^{- 1}$ along $x$ -direction and accelerates uniformly at the rate of 2.0 ${ms}^{- 2}$ . The time taken by the particle to reach the velocity of 60.0 ${ms}^{- 1}$ is ______. [2023]

6 s
3 s
30 s
25 s

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(4)

$v = u + a t$

$60 = 10 + 2 t$

$t = 25 sec$

Correct option (4)

Q 7 :

For a train engine moving with speed of 20 ${ms}^{- 1}$ , the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt{x}$ ${ms}^{- 1}$ . The value of $x$ is ______. (Assuming same retardation is produced by brakes) [2023]

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(200)

$u = 20 m/s, S_{1} = 500 m, v = 0$

By third equation of motion

$0 = {(20)}^{2} - 2 a \cdot 500 \Rightarrow a = \frac{4}{10} {m/s}^{2}$

$u = 20 m/s, S_{2} = 250 m, v = ?$

$v^{2} = {(20)}^{2} - 2 a \cdot 250 \Rightarrow v = \sqrt{200} m/s$

$\Rightarrow x = 200$

Q 8 :

A particle starts moving from time t=0 and its coordinate is given as $x (t) = 4 t^{3} - 3 t$

A. The particle returns to its original position (origin) 0.866 units later
B. The particle is 1 unit away from origin at its turning point
C. Acceleration of the particle is non-negative
D. The particle is 0.5 units away from origin at its turning point
E. Particle never turns back as acceleration is non-negative

Choose the correct answer from the options given below : [2026]

A, C, D Only
A, C Only
A, B, C Only
C, E Only

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