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NEET • PHYSICS • MOTION IN A STRAIGHT LINE

Motion in a straight line practice · Hard

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1 Hard
1 of 33
Application +4 | -1
A particle is projected with velocity v0{v_0} along x-axis.The deceleration on the particle is proportional to the square of the distance from the origin i.e., a=−ax2.a = - a{x^2}. The distance at which the particle stops is
View answer and explanation
Correct option: D
(3v022α)13{\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}
Option D ((3v022α)13{\left( {\frac{{3v_0^2}}{{2\alpha }}} \right)^{\frac{1}{3}}}) is correct because it is the distance at which the particle stops, derived by integrating the given deceleration function.
2 Hard
2 of 33
Application +4 | -1
A particle moves according to the equation dvdt=α−βv\frac{d v}{d t}=\alpha-\beta v, where α\alpha and β\beta are constants. Find velocity as a function of time. Assume, body starts from rest
View answer and explanation
Correct option: B
v=(α/β)(1−e−βt)v=(\alpha / \beta)\left(1-e^{-\beta t}\right)
Option B (v=(α/β)(1−e−βt)v=(\alpha / \beta)(1-e^{-\beta t})) is correct because it is the solution to the given differential equation with the initial condition that the body starts from rest.
3 Hard
3 of 33
Analyzing +4 | -1
A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t1{t_1} and t2{t_2} seconds respectively, then
View answer and explanation
Correct option: C
t=t1t2t = \sqrt {{t_1}{t_2}}
Option C (t=t1t2t = \sqrt {{t_1}{t_2}}) is correct. By analyzing the equations of motion for each scenario and eliminating the initial velocity, we find that the time for the dropped stone is the geometric mean of the times for the stones thrown upwards and downwards.
4 Hard
4 of 33
Application +4 | -1
A car moves from X to Y with a uniform speed vu{v_u} and returns to Y with a uniform speed vd{v_d}. The average speed for this round trip is
View answer and explanation
Correct option: A
2vdvuvd+vu\frac{{2{v_d}{v_u}}}{{{v_d} + {v_u}}}
Option A (2vdvuvd+vu\frac{{2{v_d}{v_u}}}{{{v_d} + {v_u}}}) is correct because the average speed is the total distance divided by the total time, which simplifies to the harmonic mean of the two speeds.
5 Hard
5 of 33
Analyzing +4 | -1
If position vector of a particle at time tt is r⃗=3ti^+(8−4t)j^ m\vec{r}=3 t \hat{i}+(8-4 t) \hat{j} \mathrm{~m}, then
View answer and explanation
Correct option: A
Particle is moving with constant velocity in straight line
Option A (Particle is moving with constant velocity in straight line) is correct because the position vector is a linear function of time, indicating constant velocity.
6 Hard
6 of 33
Analyzing +4 | -1
A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is 4 cm and that of iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. A similar bullet fired with the same velocity from opposite direction goes through iron first and then penetrates 2cm into wood. If a1{a_1} and a2{a_2} be the retardation offered to the bullet by wood and iron plates respectively, then
View answer and explanation
Correct option: B
a2=2a1{a_2} = 2{a_1}
Option B (a2=2a1{a_2} = 2{a_1}) is correct because the work done by the resistive forces is equal to the initial kinetic energy of the bullet. By setting up equations for both scenarios and solving for the ratio of retardations, we find that the retardation offered by the iron plate is twice that of the wooden plate.
7 Hard
7 of 33
Application +4 | -1
The velocity vector of a particle moving over a horizontal plane is given by v⃗=2ti^\vec{v}=2 t \hat{i}. The ratio of the magnitude of instantaneous acceleration of the particle at t=2sec{t}=2 \mathrm{sec} to the magnitude of displacement of the particle during the interval of first 2sec2 \mathrm{sec} is
View answer and explanation
Correct option: A
3/43/4
Option A (3/4) is correct because it is the ratio of the instantaneous acceleration to the magnitude of displacement calculated from the given velocity vector.
8 Hard
8 of 33
Application +4 | -1
The engine of a train can impart a maximum acceleration of 1ms−21\mathrm{m}{\mathrm{s}^{ - 2}} and the brakes can give a maximum retardation of 3ms−2\mathrm{3m}{\mathrm{s}^{ - 2}} . The least time during which a train can go from one place to the other place at a distance of 1.2 km is nearly
View answer and explanation
Correct option: C
56.6 s
Option C (56.6 s) is correct because it is the least time calculated considering both maximum acceleration and retardation to cover the given distance.
9 Hard
9 of 33
Application +4 | -1
The co-ordinates of a moving particle any time “tt” are given by x=αt3  &  y=βt3.x = \alpha {t^3}\,\,\& \,\,y = \beta {t^3}. The speed of the particle at time t=2  sec⁡t = 2\,\,\sec
View answer and explanation
Correct option: C
12α2+β212\sqrt {{\alpha ^2} + {\beta ^2}}
Option C (12α2+β212\sqrt {{\alpha ^2} + {\beta ^2}}) is correct because it represents the magnitude of the velocity vector at t=2s, calculated from the derivatives of x and y with respect to time.
10 Hard
10 of 33
Application +4 | -1
A particle moves 200 cm in the first 2s and 220 cm in the next 4s with uniform deceleration. The velocity of the particle at the end of 7s is
View answer and explanation
Correct option: C
10 cms−1\mathrm{cm}{\mathrm{s}^{ - 1}}
Option C (10 cms−1\mathrm{cm}{\mathrm{s}^{ - 1}}) is correct because it is the final velocity calculated using the equations of motion with uniform deceleration.
11 Hard
11 of 33
Analyzing +4 | -1
A point moves with a uniform acceleration and v1,v2,v3v_1, v_2, v_3 denote the average velocities in three successive intervals of time t1,t2,t3t_1, t_2, t_3. Which of the following relations is correct?
View answer and explanation
Correct option: B
(v1−v2):(v2−v3)=(t1+t2):(t2+t3)\left(v_1-v_2\right):\left(v_2-v_3\right)=\left(t_1+t_2\right):\left(t_2+t_3\right)
Option B ((v1−v2):(v2−v3)=(t1+t2):(t2+t3)\left(v_1-v_2\right):\left(v_2-v_3\right)=\left(t_1+t_2\right):\left(t_2+t_3\right)) is correct because it relates the differences in average velocities to the sums of the corresponding time intervals under uniform acceleration.
12 Hard
12 of 33
Application +4 | -1
The co-ordinates of a particle moving in x−yx-y plane are given by: x=2+4t,y=3t+8t2.x=2+4 t, y=3 t+8 t^2 \text {.}The motion of the particle is:
View answer and explanation
Correct option: D
Uniformly accelerated having motion along a parabolic path
Option D (Uniformly accelerated having motion along a parabolic path) is correct because the x-coordinate changes linearly with time while the y-coordinate changes quadratically, indicating uniform acceleration and a parabolic trajectory.
13 Hard
13 of 33
Application +4 | -1
If a=3t2+2ta = 3{t^2} + 2t initial velocity is 5 m/s. Find the velocity at  t=4 s\,t = 4\,s. The motion is in straight line, aa is acceleration in m/s2m/{s^2} and tt is time in seconds.
View answer and explanation
Correct option: C
85 m/s
Option C (85 m/s) is correct because it is the final velocity calculated by integrating the acceleration function and adding the initial velocity.
14 Hard
14 of 33
Application +4 | -1
The relation 3t=3x+63t = \sqrt 3 x + 6 describes the displacement of a particle in one direction, where x is in meter and t in second. The displacement when velocity is zero
View answer and explanation
Correct option: D
zero
Option D (zero) is correct because when the velocity is zero, solving for t and substituting back into the displacement equation yields zero displacement.
15 Hard
15 of 33
Application +4 | -1
A ball is dropped on the floor from a height of 10m. It rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.01 s, the average acceleration during contact is nearly [Take g=10ms−2\mathrm{g} = 10\mathrm{m}{\mathrm{s}^{ - 2}}]
View answer and explanation
Correct option: D
15002  ms−21500\sqrt 2 \mathrm{\;m}{\mathrm{s}^{ - 2}} downwards
Option C (15005  ms−21500\sqrt 5 \;\mathrm{m}{\mathrm{s}^{ - 2}} upwards) is correct because it is the average acceleration calculated from the change in velocity during the contact time.

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UNITS & MEASUREMENTS 392 MCQs WORK-ENERGY-POWER 355 MCQs MECHANICAL PROPERTIES OF FLUIDS 317 MCQs MOTION IN PLANE 307 MCQs MAGNETIC EFFECT OF CURRENT 282 MCQs THERMAL PROPERTIES OF MATTER 269 MCQs
16 Hard
16 of 33
Application +4 | -1
The driver of a car moving with a speed of 10ms−110\mathrm{m}{\mathrm{s}^{ - 1}} sees a red light ahead, applies breaks and stops after covering 10m distance. If the same car were moving with a speed of 20ms−120\mathrm{m}{\mathrm{s}^{ - 1}}, the same driver would have stopped the car after covering 30 m distance. Within what distance the car can be stopped if travelling with a velocity of 15ms−115\mathrm{m}{\mathrm{s}^{ - 1}}? Assume the same reaction time and the same deceleration in each case.
View answer and explanation
Correct option: A
18.75 m
Option A (18.75 m) is correct because it calculates the stopping distance based on the assumption of constant deceleration and reaction time.
17 Hard
17 of 33
Analyzing +4 | -1
When acceleration be function of velocity as a=f(v)a=f(v), then
View answer and explanation
Correct option: D
(1) and (3) are correct.
(1) and (3) are correct. The displacement can be found by integrating vdvf(v)\frac{v dv}{f(v)}, and the slope of the acceleration versus velocity graph can be constant.
18 Hard
18 of 33
Application +4 | -1
A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after
View answer and explanation
Correct option: A
vg[1+1+2ghv2]\frac{v}{\mathrm{g}}\left[ {1 + \sqrt {1 + \frac{{2\mathrm{g}h}}{{{v^2}}}} } \right]
Option A (vg[1+1+2ghv2]\frac{v}{\mathrm{g}}\left[ {1 + \sqrt {1 + \frac{{2\mathrm{g}h}}{{{v^2}}}} } \right]) is correct because it is derived from the equations of motion, considering the initial upward velocity and the height of the tower.
19 Hard
19 of 33
Application +4 | -1
A body moving with a uniform acceleration crosses a distance of 65 m in the 5th second and 105 m in 9th second. How far will it go in 20 s?
View answer and explanation
Correct option: C
2400 m
Option C (2400 m) is correct because it is the distance calculated based on the given information about the distance covered in the 5th and 9th seconds, assuming uniform acceleration.
20 Hard
20 of 33
Application +4 | -1
Two balls are projected upward simultaneously with speeds of 40 m/s40 {~m} / {s} and 60 m/s60 {~m} / {s}. Relative distance of the second ball with respect to the first ball at time t=5sect=5 {sec} will be [Neglect air resistance]
View answer and explanation
Correct option: C
100 m100 {~m}
Option C (100 m100 {~m}) is correct because it is the relative distance calculated using the difference in initial velocities and the time elapsed.
21 Hard
21 of 33
Application +4 | -1
A ball dropped from a point PP crosses a point QQ in tt seconds. The time taken by it to travel from Q to R, if PQ=QRPQ = QR
View answer and explanation
Correct option: D
(2−1)t\left( {\sqrt 2 - 1} \right)t
Option D ((2−1)t\left( {\sqrt 2 - 1} \right)t) is correct because it is calculated by using the equations of motion for free fall and solving for the time taken to travel from Q to R.
22 Hard
22 of 33
Application +4 | -1
A body travels 102.5 m102.5 \mathrm{~m} in nth{n}^{\text {th}} second and 115.0 m115.0 \mathrm{~m} in (n+2)th({n}+2)^{\mathrm{th}} second. The acceleration is:
View answer and explanation
Correct option: B
6.25 m/s26.25 \mathrm{~m} / \mathrm{s}^2
Option B 6.25 m/s26.25 \mathrm{~m} / \mathrm{s}^2 is correct because it is calculated using the formula for distance traveled in the nth second and solving the resulting equations.
23 Hard
23 of 33
Application +4 | -1
The displacement of a body along X- axis depends on time as x=t+1.\sqrt x = t + 1. Then the velocity of body
View answer and explanation
Correct option: A
Increases with time
Option A (Increases with time) is correct because the velocity is proportional to time, as derived from the given equation.
24 Hard
24 of 33
Application +4 | -1
A particle moving along XX-axis has acceleration ff, at time tt, given f=f0(1−tT)f=f_0\left(1-\frac{t}{T}\right), where f0f_0 and TT are constants. The particle at t=0t=0 has zero velocity. When f=0f=0, the particle's velocity (vx)\left(v_x\right) is
View answer and explanation
Correct option: A
12f0T\frac{1}{2} f_0 T
Option A (12f0T\frac{1}{2} f_0 T) is correct because it is derived by integrating the acceleration function and using the condition that f=0.
25 Hard
25 of 33
Application +4 | -1
Two particles are moving with velocities 22 m/sec and
3 m/sec along X-axis in opposite direction. Their initial
positions are at x = 30 m, x = 180 m respectively. find
time interval after which they meet with each other
View answer and explanation
Correct option: C
6 sec
Option C (6 sec) is correct because it is the time calculated by considering the relative velocity and the initial separation between the two particles.
26 Hard
26 of 33
Application +4 | -1
Amongst the following equation of motion, which represents uniformly accelerated motion?
View answer and explanation
Correct option: C
t=x+abt=\sqrt{\frac{x+a}{b}}
Option C (t=x+abt=\sqrt{\frac{{x+a}}{{b}}}) is correct because solving for x yields an equation of the form x=bt2−ax = bt^2 - a, which represents uniformly accelerated motion.
27 Hard
27 of 33
Analyzing +4 | -1
The distance x covered by a particle in one-dimensional motion varies with time t as x2=at2+2bt+c{x^2} = a{t^2} + 2bt + c. The accelerating of the particle varies as
View answer and explanation
Correct option: A
x−3{x^{ - 3}}
Option A (x−3{x^{ - 3}}) is correct because the acceleration is inversely proportional to the cube of the distance x, derived from differentiating the given equation.
28 Hard
28 of 33
Application +4 | -1
Two bodies move in a straight line towards each other with initial velocities v1v_1 and v2v_2 and with constant accelerations a1a_1 and a2a_2 directed against their corresponding initial velocities. The maximum initial separation ℓmax⁡\ell_{\max } between them for which they will meet during the motion is:
View answer and explanation
Correct option: B
(v1+v2)22(a1+a2)\frac{\left(v_1+v_2\right)^2}{2\left(a_1+a_2\right)}
Option B ((v1+v2)22(a1+a2)\frac{\left(v_1+v_2\right)^2}{2\left(a_1+a_2\right)}) is correct because it represents the maximum initial separation for the two bodies to meet, considering their initial velocities and decelerations.
29 Hard
29 of 33
Application +4 | -1
A ball is thrown vertically upwards. It was observed at a height h twice with a time interval Δ\Deltat. The initial velocity of the ball is
View answer and explanation
Correct option: C
128gh+g2(Δt)2\frac{1}{2}\sqrt {8\mathrm{g}h + {\mathrm{g}^2}{{\left( {\Delta t} \right)}^2}}
Option C (128gh+g2(Δt)2\frac{1}{2}\sqrt {8\mathrm{g}h + {\mathrm{g}^2}{{\left( {\Delta t} \right)}^2}}) is correct because it is derived from the equations of motion, considering the time interval between the ball passing the height h twice.
30 Hard
30 of 33
Understanding +4 | -1
Assertion: The displacement-time graph of a body moving with uniform acceleration is a parabola.
Reason: The displacement is proportional to time for uniformly accelerated motion.
View answer and explanation
Correct option: C
If Assertion is true but Reason is false
Option C is correct because the assertion is true as displacement-time graph for uniform acceleration is a parabola, but the reason is false because displacement is proportional to the square of time.
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