Motion in One Dimension (Straight Line) – High Difficulty MCQ Worksheet

Level: JEE Main / Advanced, NEET (Higher Order Thinking)

Q1.

A particle starts from rest with acceleration $a=6t-2\ \text{m s}^{-2}$. The velocity becomes zero again at

A. $\frac{2}{3}\,\text{s}$

B. $\frac{4}{3}\,\text{s}$

C. $2\text{s}$

D. Never

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Q2.

A particle moves along the x-axis according to

$$x=t^3-6t^2+9t+4$$

The particle changes its direction of motion

A. Once

B. Twice

C. Three times

D. Never

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Q3.

A car moving at $20\,\text{m/s}$ accelerates uniformly at $2\,\text{m/s}^2$ for 5 s and then decelerates uniformly at $5\,\text{m/s}^2$ until it stops. The total distance travelled is

A. 170 m

B. 180 m

C. 215 m

D. 235 m

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Q4.

A particle moves such that

$$v=(x-2)^2-4$$

At which position is acceleration zero (excluding turning points)?

A. $x=2$

B. $x=0$

C. $x=4$

D. None

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Q5.

Two particles start simultaneously from the same point.

$$x_1=3t+t^2,\qquad x_2=2t+2t^2$$

The first instant when they meet again is

A. 1 s

B. 2 s

C. 3 s

D. Never

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Q6.

A particle moves with

$$v=4t-3$$

The displacement during the third second is

A. 7 m

B. 9 m

C. 12 m

D. 13 m

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Q7.

A body is projected vertically upward with speed $40\,\text{m/s}$. Taking $g=10\,\text{m/s}^2$, the ratio of distances covered during the first and second seconds is

A. 7:5

B. 5:3

C. 3:1

D. 9:7

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Q8.

A particle moves according to

$$x=4t^3-12t^2+9t$$

The maximum speed during the interval $0\le t\le2$ s is

A. 3 m/s

B. 12 m/s

C. 9 m/s

D. 15 m/s

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Q9.

A train moving at $25\,\text{m/s}$ applies brakes producing retardation of $2.5\,\text{m/s}^2$. The distance covered during the last 4 seconds before stopping is

A. 20 m

B. 25 m

C. 30 m

D. 40 m

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Q10.

The velocity-time graph is a straight line joining (0,4) and (8,-4). The total distance travelled is

A. 12 m

B. 16 m

C. 20 m

D. 24 m

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Q11.

A particle moves with acceleration

$$a=-4v$$

Initially $v=10\,\text{m/s}$. The velocity after $0.25$ s is approximately

A. $3.68\,\text{m/s}$

B. $5\,\text{m/s}$

C. $7\,\text{m/s}$

D. $1\,\text{m/s}$

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Q12.

The displacement of a particle is

$$x=5+8t-3t^2$$

The total distance travelled before changing direction is

A. 4 m

B. $\frac{16}{3}$ m

C. 6 m

D. 8 m

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Q13.

A particle starts from rest with constant jerk (rate of change of acceleration)

$$j=2\,\text{m/s}^3$$

The distance travelled in 3 s is

A. 6 m

B. 9 m

C. 12 m

D. 18 m

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Q14.

A body covers half the distance with speed $v$ and the remaining half with speed $3v$. Its average speed is

A. $2v$

B. $\frac{3v}{2}$

C. $\frac{3v}{4}$

D. $\frac{6v}{4}$

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Q15.

A particle moves such that

$$v^2=16-2x$$

The retardation is

A. $1\,\text{m/s}^2$

B. $2\,\text{m/s}^2$

C. $4\,\text{m/s}^2$

D. Variable

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Q16.

Two cars move toward each other with speeds 15 m/s and 20 m/s, initially 350 m apart. They continue uniformly. Meeting time is

A. 8 s

B. 9 s

C. 10 s

D. 12 s

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Q17.

A particle moves according to

$$x=2\sin(\pi t)$$

Its maximum speed equals

A. $2\pi$

B. $\pi$

C. $4\pi$

D. $8\pi$

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Q18.

A stone is dropped from a balloon rising vertically at $10\,\text{m/s}$. It reaches the ground after 5 s. Height of release is

A. 75 m

B. 100 m

C. 125 m

D. 150 m

(Take $g=10\,\text{m/s}^2$.)

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Q19.

A particle has

$$x=t^4-8t^2$$

The number of instants at which acceleration becomes zero is

A. 0

B. 1

C. 2

D. 4

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Q20.

The position of a particle is

$$x=6t-3t^2+t^3$$

The maximum displacement from the origin before $t=4$ s is

A. 4 m

B. 6 m

C. 8 m

D. 9 m

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Answer Key

Q	Ans	Q	Ans
1	B	11	A
2	B	12	B
3	C	13	B
4	A	14	B
5	A	15	A
6	A	16	C
7	A	17	A
8	C	18	A
9	A	19	B
10	B	20	C

These questions are intentionally designed to be higher than standard NCERT level, with a mix of calculus-based kinematics, graphical reasoning, relative motion, variable acceleration, jerk, stopping distance, and displacement analysis suitable for advanced competitive exam practice.

Motion in One Dimension (Straight Line)

Hints for Solving All 20 Questions (JEE Main/Advanced Level)

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Q1 Hint

• Integrate the acceleration function to obtain velocity.
• Use the initial condition $v=0$ at $t=0$.
• Equate the velocity expression to zero and solve for $t$.

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Q2 Hint

• Differentiate displacement to obtain velocity.
• Factorize the velocity equation.
• Identify where velocity changes sign.
• The number of sign changes equals the number of direction changes.

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Q3 Hint

Break the motion into two parts:

1. Acceleration phase
2. Braking phase

Use

$$v=u+at$$

for the first part and

$$v^2=u^2+2as$$

for the second.

Finally,

$$s_{\text{total}}=s_1+s_2$$

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Q4 Hint

Use

$$a=v\frac{dv}{dx}$$

Differentiate $v(x)$.

Acceleration becomes zero when either

• $v=0$, or
• $\frac{dv}{dx}=0$
  

Read the question carefully to know which one is required.

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Q5 Hint

Equate the two displacement equations.

$$x_1=x_2$$

Solve for $t$.

Ignore the trivial solution $t=0$.

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Q6 Hint

Displacement during the third second is

$$x(3)-x(2)$$

Integrate velocity first to obtain displacement.

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Q7 Hint

Use Galileo's Law of Odd Numbers.

Distance covered in successive seconds during upward motion follows

$$u-\frac{g}{2}(2n-1)$$

or simply calculate using

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

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Q8 Hint

Differentiate displacement to obtain velocity.

Find

• critical points where acceleration becomes zero,
• endpoints of the interval.

Compare the magnitude of velocity.

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Q9 Hint

Find stopping time.

The "last four seconds" means consider motion between

$$t=t_{\text{stop}}-4$$

and

$$t=t_{\text{stop}}$$

Use displacement equations.

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Q10 Hint

Distance equals the area under the speed-time graph.

Split the graph wherever velocity changes sign.

Add absolute areas.

Do not calculate displacement.

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Q11 Hint

Recognize the differential equation

$$\frac{dv}{dt}=-4v$$

Solve using

$$v=v_0e^{-4t}$$

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Q12 Hint

Find when velocity becomes zero.

That is the turning point.

Calculate displacement till that instant.

Since motion does not reverse before then,

distance = displacement.

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Q13 Hint

Remember

$$j=\frac{da}{dt}$$

Integrate successively

Jerk → Acceleration

Acceleration → Velocity

Velocity → Displacement

Apply initial conditions.

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Q14 Hint

Equal distances imply

Average speed

$$=\frac{2v_1v_2}{v_1+v_2}$$

No need to calculate time separately.

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Q15 Hint

Use

$$a=v\frac{dv}{dx}$$

Differentiate

$$v^2=16-2x$$

instead of taking square roots.

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Q16 Hint

Use relative speed.

Since they move toward each other,

$$v_{\text{relative}} =v_1+v_2$$

Then

$$t=\frac{\text{distance}}{\text{relative speed}}$$

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Q17 Hint

Differentiate displacement.

Maximum speed occurs when

$$|\cos(\pi t)|=1$$

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Q18 Hint

Initial velocity is upward because the balloon is moving upward.

Use

$$h=ut+\frac12at^2$$

Take downward as positive or upward as positive—but remain consistent.

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Q19 Hint

Differentiate twice.

Acceleration

$$a=\frac{d^2x}{dt^2}$$

Set acceleration equal to zero.

Count valid real values of $t$.

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Q20 Hint

Differentiate displacement.

Velocity

$$v=\frac{dx}{dt}$$

Find turning points.

Evaluate displacement at

• turning points
• interval endpoints

Choose the maximum.

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Useful Formula Sheet

Constant Acceleration

$$v=u+at$$
$$s=ut+\frac12at^2$$
$$v^2=u^2+2as$$
$$s=\frac{(u+v)}{2}t$$

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Variable Acceleration

$$a=\frac{dv}{dt}$$ $$a=v\frac{dv}{dx}$$

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Jerk

$$j=\frac{da}{dt}$$

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Average Speed

$$v_{\text{avg}} =\frac{\text{Total Distance}}{\text{Total Time}}$$

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Equal Distance Average Speed

$$v_{\text{avg}} = \frac{2v_1v_2}{v_1+v_2}$$

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Relative Speed

Opposite directions:

$$v_r=v_1+v_2$$

Same direction:

$$v_r=|v_1-v_2|$$

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Distance in $n^{\text{th}}$ Second

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

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Graph Facts

• Area under $v$-$t$ graph → Displacement
• Area under speed–time graph → Distance
• Slope of $x$-$t$ graph → Velocity
• Slope of $v$-$t$ graph → Acceleration

These hints are designed to guide students toward the solution without revealing the complete calculations, making them suitable for self-practice at the JEE Main/Advanced and NEET level.