Motion in a Straight Line — JEE Main practice

15 questions

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Sample questions with solutions

Q1 · 2026

A paratrooper jumps from an aeroplane and opens a parachute after $2\ \mathrm{s}$ of free fall and starts deaccelerating with $3\ \mathrm{m/s^2}$ . At $10\ \mathrm{m}$ height from ground, while descending with the help of parachute, the speed of paratrooper is $5\ \mathrm{m/s}$ . The initial height of the airplane is ____ m.

\left(g=10\ \mathrm{m/s^2}\right)

A.
$92.5$
B.
$62.5$
C.
$20$
D.
$82.5$

Answer: A

Divide the motion into two intervals.
- Interval 1: Free fall for $2\ \mathrm{s}$ .
- Interval 2: Motion after the parachute opens with a constant deacceleration of $3\ \mathrm{m/s^2}$ .
Interval 1: Free fall

The paratrooper starts from rest:
$u_1=0,\qquad t_1=2\ \mathrm{s}$
Distance covered:
$h_1=u_1t_1+\frac{1}{2}gt_1^2$ $=0+\frac{1}{2}\times10\times2^2 =20\ \mathrm{m}$
Velocity at the end of free fall:
$v_1=u_1+gt_1$ $=0+10\times2 =20\ \mathrm{m/s}$
Interval 2: After opening the parachute

Initial velocity:
$u_2=20\ \mathrm{m/s}$
Final velocity at a height of $10\ \mathrm{m}$ above the ground:
$v_2=5\ \mathrm{m/s}$
Acceleration:
$a=-3\ \mathrm{m/s^2}$
Using the third equation of motion:
$v_2^2=u_2^2+2ah_2$
Substituting the values:
$5^2=20^2+2(-3)h_2$ $25=400-6h_2$ $h_2=\frac{375}{6}=62.5\ \mathrm{m}$
The total initial height of the airplane is:
$H=h_1+h_2+10$ $=20+62.5+10 =92.5\ \mathrm{m}$
Therefore, the airplane was initially at a height of $92.5\ \mathrm{m}$ , so the correct option is A.

Q2 · 2026

The velocity $(v)$ – Distance $(x)$ graph is shown in figure. Which graph represents acceleration $(a)$ versus distance $(x)$ variation of this system?

Answer: D

The given $v$ – $x$ graph is a straight line with a negative slope and a positive intercept.

Hence, it can be written as:
$v=-m𝑥+v_0$
where $m>0$ .
The relation between acceleration and velocity is:
$a=v\frac{dv}{dx}$
Differentiate the velocity equation:
$\frac{dv}{dx}=-m$
Substitute into the acceleration formula:
$a=(-m𝑥+v_0)(-m)$ $a=m^2𝑥-mv_0$
This is the equation of a straight line.
- Slope:
  $m^2>0$
  so the graph has a positive slope.
- Intercept:
  $-mv_0<0$
  so the graph cuts the acceleration axis below the origin.
Therefore, the correct acceleration–distance graph is the straight line with positive slope and negative intercept, which corresponds to Option D.

Q3 · 2026

Water drops fall from a tap on the floor, $5\ \mathrm{m}$ below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is ____ m.

\left(g=10\ \mathrm{m/s^2}\right)

A.
$3.8$
B.
$4.0$
C.
$4.2$
D.
$2.5$

Answer: C

The first drop falls from rest through a height of:
$H=5\ \mathrm{m}$
Using the equation of motion,
$s=ut+\frac{1}{2}gt^2$
Since $u=0$ ,
$5=\frac{1}{2}\times10\times T^2$ $5=5T^2$ $T=1\ \mathrm{s}$
The first drop reaches the ground exactly when the sixth drop starts falling.

Therefore, the total time of $1\ \mathrm{s}$ is divided into five equal intervals:
$\Delta t=\frac{1}{5}=0.2\ \mathrm{s}$
At the instant the first drop hits the ground:
- Drop 1 has fallen for $1.0\ \mathrm{s}$ .
- Drop 2 has fallen for $0.8\ \mathrm{s}$ .
- Drop 3 has fallen for $0.6\ \mathrm{s}$ .
- Drop 4 has fallen for $0.4\ \mathrm{s}$ .
The distance fallen by the fourth drop is:
$h_4=\frac{1}{2}gt^2$ $=\frac{1}{2}\times10\times(0.4)^2$ $=5\times0.16 =0.8\ \mathrm{m}$
Hence, its height above the ground is:
$H-h_4=5-0.8=4.2\ \mathrm{m}$
Therefore, the height of the fourth drop from the ground is $4.2\ \mathrm{m}$ , so the correct option is C.

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