A particle starts moving from time t = 0$t = 0$ and its coordinate is given as x(t) = 4t^3 - 3t$x(t) = 4t^{3} - 3t$ .
A. The particle returns to its original position (origin) 0.866$0.866$ 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:
A.textA,C,D only$\text{A,C,D only}$
B.textA,B,C only$\text{A,B,C only}$
C.textC,E only$\text{C,E only}$
D.textA,C only$\text{A,C only}$
Solution & Explanation
### Related Formula
v = fracdxdt$$v = \frac{dx}{dt}$$a = fracdvdt$$a = \frac{dv}{dt}$$
### Core Logic
Given x(t) = 4t^3 - 3t$x(t) = 4t^3 - 3t$.
The particle returns to origin when x = 0$x = 0$:
4t^3 - 3t = 0 Rightarrow t(4t^2 - 3) = 0$$4t^3 - 3t = 0 \Rightarrow t(4t^2 - 3) = 0$$t = 0$t = 0$ (start) and t = sqrtfrac34 = fracsqrt32 approx 0.866 text s$t = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.866 \text{ s}$. Thus, statement A is correct.
### Step 1: Velocity and Turning Point
Velocity is v = fracdxdt = 12t^2 - 3$v = \frac{dx}{dt} = 12t^2 - 3$.
At turning point, velocity becomes zero:
12t^2 - 3 = 0 Rightarrow t^2 = frac14 Rightarrow t = frac12 text s$$12t^2 - 3 = 0 \Rightarrow t^2 = \frac{1}{4} \Rightarrow t = \frac{1}{2} \text{ s}$$
(We take t > 0$t > 0$ since motion starts at t=0$t=0$).
### Step 2: Position at Turning Point
Substitute t = 1/2$t = 1/2$ into x(t)$x(t)$:
xleft(frac12right) = 4left(frac18right) - 3left(frac12right) = frac12 - frac32 = -1$$x\left(\frac{1}{2}\right) = 4\left(\frac{1}{8}\right) - 3\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{3}{2} = -1$$
The particle is |-1| = 1$|-1| = 1$ unit away from origin. Statement B is correct, and D is incorrect.
### Step 3: Acceleration check
Acceleration a = fracdvdt = 24t$a = \frac{dv}{dt} = 24t$.
Since t geq 0$t \geq 0$, a geq 0$a \geq 0$, so acceleration is always non-negative. Statement C is correct.
Statement E is incorrect because the particle does turn back (at t = 0.5$t = 0.5$, velocity changes sign from negative to positive).
### Step 4: Final Conclusion
Statements A, B, and C are correct.
### Pattern Recognition
To analyze 1D motion, simply find roots of x(t)=0$x(t)=0$ (origin passes), v(t)=0$v(t)=0$ (turning points), and check the sign of a(t)$a(t)$ over the given domain t geq 0$t \geq 0$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
Keywords:#coordinate is given as x(t)#JEE Main 2026 Evening Q33#Kinematics JEE Main 2026#Motion in a Straight Line JEE Main 2026
More Kinematics Previous-Year Questions
Q39jee_main_2026_21_jan_eveningRelative Motion in 2D
A river of width 200 text m$200 \text{ m}$ is flowing from west to east with a speed of 18 text km/h$18 \text{ km/h}$. A boat, moving with speed of 36 text km/h$36 \text{ km/h}$ in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ________ and ________ respectively.
A.20 text s and 100 text m$20 \text{ s and } 100 \text{ m}$
B.40 text s and 0 text m$40 \text{ s and } 0 \text{ m}$
C.40 text s and 200 text m$40 \text{ s and } 200 \text{ m}$
D.40 text s and 100 text m$40 \text{ s and } 100 \text{ m}$
Solution
### Related Formula
t_textmin = fracdv_br$$t_{\text{min}} = \frac{d}{v_{br}}$$textDrift (displacement along bank) = v_r times t_texttotal$$\text{Drift (displacement along bank)} = v_r \times t_{\text{total}}$$
### Core Logic
First, convert speeds to m/s.
River speed v_R = 18 times frac518 = 5 text m/s$v_R = 18 \times \frac{5}{18} = 5 \text{ m/s}$.
Boat speed in still water v_BR = 36 times frac518 = 10 text m/s$v_{BR} = 36 \times \frac{5}{18} = 10 \text{ m/s}$.
River width d = 200 text m$d = 200 \text{ m}$.
River boat vector diagram for Q39 - JEE Main 2026 Evening
For minimum time across the river, the boat must head exactly perpendicular to the river flow.
### Step 1: Calculating Minimum Time
Time taken for one crossing (bank to bank):
t_textone-way = fracdv_BR = frac20010 = 20 text s$$t_{\text{one-way}} = \frac{d}{v_{BR}} = \frac{200}{10} = 20 \text{ s}$$
Since it is a round trip (bank to bank and back), total minimum time is:
t_texttotal = 2 times 20 = 40 text s$$t_{\text{total}} = 2 \times 20 = 40 \text{ s}$$
### Step 2: Calculating Displacement along Bank
Displacement along the river bank (drift) happens exclusively due to the river's flow during this total time:
textDisplacement = v_R times t_texttotal = 5 times 40 = 200 text m$$\text{Displacement} = v_R \times t_{\text{total}} = 5 \times 40 = 200 \text{ m}$$
### Pattern Recognition
Minimum time always ignores drift and maxes out the perpendicular velocity vector. Drift simply becomes V_textriver times T_texttotal$V_{\text{river}} \times T_{\text{total}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
Q30jee_main_2026_23_january_eveningMotion under Gravity
A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 \, mathrmm/s^2$3 \, \mathrm{m/s}^{2}$ . At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the aeroplane is ____ m.
(g = 10 m/s ^2$^{2}$ )
A.62.5$62.5$
B.92.5$92.5$
C.20$20$
D.82.5$82.5$
Solution
### Related Formula
s = ut + frac12at^2$$s = ut + \frac{1}{2}at^2$$v = u + at$v = u + at$v^2 = u^2 + 2as$v^2 = u^2 + 2as$
### Core Logic
The motion is divided into three distinct vertical sections:
Motion under Gravity diagram for Q30 - JEE Main 2026 EveningMotion under Gravity diagram for Q30 - JEE Main 2026 Evening
1) Free fall phase (A to B$A \to B$)
2) Deceleration phase with parachute (B to C$B \to C$)
3) Final descent phase to the ground (C to D$C \to D$)
### Step 1: Free Fall (A to B)
Initial velocity u = 0$u = 0$, t = 2 \, mathrms$t = 2 \, \mathrm{s}$, g = 10 \, mathrmm/s^2$g = 10 \, \mathrm{m/s^2}$.
x_1 = frac12 times 10 times 2^2 = 20 \, mathrmm$$x_1 = \frac{1}{2} \times 10 \times 2^2 = 20 \, \mathrm{m}$$
Velocity at B:
V_B = 0 + 10 times 2 = 20 \, mathrmm/s$$V_B = 0 + 10 \times 2 = 20 \, \mathrm{m/s}$$
### Step 2: Deceleration Phase (B to C)
Starts with V_B = 20 \, mathrmm/s$V_B = 20 \, \mathrm{m/s}$, decelerates at 3 \, mathrmm/s^2$3 \, \mathrm{m/s^2}$ (a = -3 \, mathrmm/s^2$a = -3 \, \mathrm{m/s^2}$), and reaches V_C = 5 \, mathrmm/s$V_C = 5 \, \mathrm{m/s}$.
V_C^2 = V_B^2 + 2a x_2$$V_C^2 = V_B^2 + 2a x_2$$5^2 = 20^2 - 2(3)x_2$$5^2 = 20^2 - 2(3)x_2$$25 = 400 - 6x_2$25 = 400 - 6x_2$6x_2 = 375 implies x_2 = frac3756 = 62.5 \, mathrmm$$6x_2 = 375 \implies x_2 = \frac{375}{6} = 62.5 \, \mathrm{m}$$
### Step 3: Final Phase and Total Height
The last phase x_3$x_3$ is given as 10 \, mathrmm$10 \, \mathrm{m}$ above the ground.
H = x_1 + x_2 + x_3$$H = x_1 + x_2 + x_3$$H = 20 + 62.5 + 10 = 92.5 \, mathrmm$$H = 20 + 62.5 + 10 = 92.5 \, \mathrm{m}$$
### Pattern Recognition
Split multi-stage motion problems into clear intervals. The final velocity of interval N becomes the initial velocity of interval N+1. Calculate displacement for each interval independently and sum them up.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
Q44jee_main_2026_24_january_eveningMotion Graphs
The velocity (v) - Distance (x) graph is shown in figure. Which graph represents acceleration (a) versus distance (x) variation of this system?
A linear descending velocity versus distance graph.
A. Option 1
B. Option 2
C. Option 3
D. Option 4
Solution
### Related Formula
a = v fracdvdx$$a = v \frac{dv}{dx}$$
### Core Logic
A linear descending velocity versus distance graph.
Equation of V$V$ vs x$x$ from the provided graph is a straight line with a negative slope:
V = C_1 - C_2 x$V = C_1 - C_2 x$
(where C_1, C_2 > 0$C_1, C_2 > 0$).
### Step 1: Calculate Acceleration
Differentiate V$V$ with respect to x$x$:
fracdVdx = -C_2$$\frac{dV}{dx} = -C_2$$
Substitute into the acceleration formula:
a = (C_1 - C_2 x) times (-C_2)$$a = (C_1 - C_2 x) \times (-C_2)$$a = C_2^2 x - C_1 C_2$$a = C_2^2 x - C_1 C_2$$
### Step 2: Analyze Graph Profile
The equation a = C_2^2 x - C_1 C_2$a = C_2^2 x - C_1 C_2$ represents a straight line with a **positive slope** (C_2^2$C_2^2$) and a **negative y-intercept** (-C_1 C_2$-C_1 C_2$).
A linear descending velocity versus distance graph.
Therefore, the graph is a straight line intercepting the negative a-axis and moving positively.
### Pattern Recognition
A linear negative v-x$v-x$ graph always yields an a-x$a-x$ graph which is linear with a positive slope and negative intercept. Memorizing a = v(dv/dx)$a = v(dv/dx)$ maps graphical traits instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
Qjee_main_2025_02_april_morningMotion in Two Dimensions
A river is flowing from west to east direction with speed of 9mathrm~km~h^-1$9\mathrm{~km~h}^{-1}$. If a boat capable of moving at a maximum speed of 27mathrm~km~h^-1$27\mathrm{~km~h}^{-1}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of 150^circ$150^{\circ}$ to direction of river flow, then the width of the river is:
### Related Formula
v_perp = v_br sintheta$$v_{\perp} = v_{br} \sin\theta$$w = v_perp cdot t$$w = v_{\perp} \cdot t$$
### Core Logic
Let the West-to-East river flow direction be along the positive x$x$-axis.
The boat's velocity relative to water is v_br = 27mathrm~km/h$v_{br} = 27\mathrm{~km/h}$ at an angle of 150^circ$150^{\circ}$ with respect to the river flow (which is 30^circ$30^{\circ}$ upstream from the perpendicular crossing line).
To find the width of the river, we only need the component of the boat's velocity perpendicular to the river bank (y$y$-axis):
v_perp = v_br sin(150^circ) = 27 sin(150^circ) = 27 times frac12 = 13.5mathrm~km/h$$v_{\perp} = v_{br} \sin(150^{\circ}) = 27 \sin(150^{\circ}) = 27 \times \frac{1}{2} = 13.5\mathrm{~km/h}$$
Convert this velocity to SI units:
v_perp = 13.5 times frac518mathrm~m/s = 3.75mathrm~m/s$$v_{\perp} = 13.5 \times \frac{5}{18}\mathrm{~m/s} = 3.75\mathrm{~m/s}$$
Given the crossing time is half a minute (t = 30mathrm~s$t = 30\mathrm{~s}$):
w = v_perp cdot t = 3.75 times 30 = 112.5mathrm~m$$w = v_{\perp} \cdot t = 3.75 \times 30 = 112.5\mathrm{~m}$$
### Step 1: Final Conclusion
The width of the river is 112.5mathrm~m$112.5\mathrm{~m}$.
### Pattern Recognition
In river-boat crossing problems, river velocity (v_r$v_r$) only causes drift along the bank; it has zero impact on crossing time or the crossing width calculations when the angle is defined relative to the flow. Use the vertical velocity component exclusively.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
More Kinematics Questions — jee_main_2026_28_january_evening
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