Online Test — Motion in a Straight Line
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
Displacement of a particle in straight-line motion is a:
Scalar
Vector
Tensor
Dimensionless number
Explanation: Displacement has magnitude and direction (a sign in 1D), so it is a vector.
Question 2 of 20
A particle moves from x = 3 m to x = -5 m. Its displacement is:
2 m
8 m
-8 m
-2 m
Explanation: $\Delta x = -5 - 3 = -8\,\text{m}$, directed along $-x$.
Question 3 of 20
Path length compared with the magnitude of displacement is always:
Smaller
Equal
Greater than or equal
Zero
Explanation: Path length can never be less than the size of the straight-line displacement.
Question 4 of 20
Instantaneous velocity is defined as:
$\frac{\Delta x}{\Delta t}$ for large $\Delta t$
$\frac{dx}{dt}$
$\frac{dv}{dt}$
$\frac{d^2x}{dt^2}$
Explanation: Instantaneous velocity is the time derivative of position, $v = \dfrac{dx}{dt}$.
Question 5 of 20
The slope of a velocity-time graph gives:
Displacement
Acceleration
Speed
Jerk
Explanation: Slope of the $v$-$t$ graph is $\dfrac{dv}{dt} = a$, the acceleration.
Question 6 of 20
For x = 5t^2 (SI), the acceleration is:
5 m/s$^2$
10 m/s$^2$
2.5 m/s$^2$
Zero
Explanation: $v = 10t$ and $a = \dfrac{dv}{dt} = 10\,\text{m/s}^2$ (constant).
Question 7 of 20
A car accelerates from 4 m/s to 28 m/s in 8 s. Its acceleration is:
2 m/s$^2$
3 m/s$^2$
4 m/s$^2$
6 m/s$^2$
Explanation: $a = \dfrac{v - u}{t} = \dfrac{28 - 4}{8} = 3\,\text{m/s}^2$.
Question 8 of 20
36 km/h equals:
5 m/s
10 m/s
18 m/s
100 m/s
Explanation: $36 \times \dfrac{5}{18} = 10\,\text{m/s}$.
Question 9 of 20
A body starts from rest with a = 4 m/s$^2$. Distance covered in 5 s is:
20 m
40 m
50 m
100 m
Explanation: $s = ut + \dfrac{1}{2}at^2 = 0 + \dfrac{1}{2}(4)(25) = 50\,\text{m}$.
Question 10 of 20
Which equation does not contain time t?
$v = u + at$
$s = ut + \frac{1}{2}at^2$
$v^2 = u^2 + 2as$
$s = \frac{(u+v)}{2}t$
Explanation: $v^2 = u^2 + 2as$ relates velocities, acceleration and displacement without $t$.
Question 11 of 20
A stone dropped from rest falls for 2 s. Its speed on hitting the ground is (g = 10 m/s$^2$):
10 m/s
20 m/s
40 m/s
5 m/s
Explanation: $v = u + gt = 0 + 10 \times 2 = 20\,\text{m/s}$.
Question 12 of 20
A ball thrown up with 40 m/s reaches a maximum height of (g = 10 m/s$^2$):
40 m
80 m
120 m
160 m
Explanation: $H = \dfrac{u^2}{2g} = \dfrac{40^2}{2 \times 10} = \dfrac{1600}{20} = 80\,\text{m}$.
Question 13 of 20
The area under a velocity-time graph between two instants gives the:
Acceleration
Displacement
Average speed
Jerk
Explanation: Area under the $v$-$t$ curve equals the displacement over that interval.
Question 14 of 20
A body has positive velocity and negative acceleration. It is:
Speeding up
At rest
Slowing down
Moving uniformly
Explanation: Opposite signs of $v$ and $a$ mean retardation — the body slows down.
Question 15 of 20
Two cars move toward each other at 12 m/s and 8 m/s. Their relative speed of approach is:
4 m/s
12 m/s
20 m/s
8 m/s
Explanation: For opposite directions, speeds add: $12 + 8 = 20\,\text{m/s}$.
Question 16 of 20
A position-time graph that is a straight line sloping upward represents:
Rest
Uniform velocity
Uniform acceleration
Retardation
Explanation: Constant slope of the $x$-$t$ graph means constant (uniform) velocity.
Question 17 of 20
The acceleration due to gravity for a freely falling body is:
Different for heavier bodies
Greater for lighter bodies
The same for all bodies
Zero
Explanation: Ignoring air resistance, all bodies fall with the same $g$ regardless of mass.
Question 18 of 20
A train slows from 30 m/s to rest in 250 m. Its retardation is:
1.0 m/s$^2$
1.8 m/s$^2$
2.0 m/s$^2$
3.6 m/s$^2$
Explanation: $v^2 = u^2 + 2as \Rightarrow 0 = 900 + 2a(250) \Rightarrow a = -1.8\,\text{m/s}^2$.
Question 19 of 20
The distance covered by a body starting from rest in the n-th second is proportional to:
$n$
$n^2$
$(2n - 1)$
$\sqrt{n}$
Explanation: $s_n = u + \dfrac{a}{2}(2n - 1)$; with $u = 0$ it is proportional to $(2n - 1)$.
Question 20 of 20
A v-t graph is a horizontal line above the time axis. This means the body has:
Zero velocity
Constant non-zero velocity
Constant acceleration
Increasing acceleration
Explanation: A horizontal $v$-$t$ line means velocity is constant and acceleration (its slope) is zero.