iCON Education - Kinematics-1D Advanced Test

Ram takes path 1 (straight line) to go from P to Q and Shyam takes path 2 (semicircle) from the image given below.
(a) Find the distance travelled by Ram and Shyam.
(b) Find the displacement of Ram and Shyam.

(a) Ram: 100 m, Shyam: 100 m; (b) Ram: 100 m, Shyam: 50 $\pi$ m

(a) Ram: 100 m, Shyam: 50 $\pi$ m; (b) Ram: 100 m, Shyam: 100 m

(a) Ram: 50 $\pi$ m, Shyam: 100 m; (b) Ram: 50 $\pi$ m, Shyam: 100 m

(a) Ram: 100 m, Shyam: 100 $\pi$ m; (b) Ram: 50 m, Shyam: 100 m

The position $x$ (in metre) of a particle varies with time $t$ (in second) as $t = \sqrt{x} + 3$. Calculate the
(a) displacement of the particle from $t = 0$ to $t = 3\text{ s}$
(b) displacement of the particle from $t = 3$ to $t = 6\text{ s}$
(c) distance travelled and displacement from $t = 0$ to $t = 6\text{ s}$

(a) $-9\text{ m}$; (b) $+9\text{ m}$; (c) Distance: $18\text{ m}$, Displacement: $0\text{ m}$

(a) $+9\text{ m}$; (b) $-9\text{ m}$; (c) Distance: $0\text{ m}$, Displacement: $18\text{ m}$

(a) $-9\text{ m}$; (b) $+9\text{ m}$; (c) Distance: $0\text{ m}$, Displacement: $0\text{ m}$

(a) $+9\text{ m}$; (b) $+9\text{ m}$; (c) Distance: $18\text{ m}$, Displacement: $18\text{ m}$

Calculate the average speed and the average velocity for a train that travels from one station to another at a uniform speed of 40 kmh$^{-1}$ and returns to the first station at a speed of 60 kmh$^{-1}$.

Average Speed: 48 kmh$^{-1}$, Average Velocity: 0 kmh$^{-1}$

Average Speed: 50 kmh$^{-1}$, Average Velocity: 10 kmh$^{-1}$

Average Speed: 48 kmh$^{-1}$, Average Velocity: 48 kmh$^{-1}$

Average Speed: 50 kmh$^{-1}$, Average Velocity: 0 kmh$^{-1}$

Calculate the average speed and the average velocity for a man who walks at a speed of 1 ms$^{-1}$ for the first one minute and then runs at a speed of 3 ms$^{-1}$ for the next one minute along a straight track.

Average Speed: 1.5 ms$^{-1}$, Average Velocity: 1.5 ms$^{-1}$

Average Speed: 2 ms$^{-1}$, Average Velocity: 2 ms$^{-1}$

Average Speed: 2 ms$^{-1}$, Average Velocity: 0 ms$^{-1}$

Average Speed: 3 ms$^{-1}$, Average Velocity: 1 ms$^{-1}$

Calculate the average speed and the average velocity for a man who walks 720 m at a uniform speed of 2 ms$^{-1}$, then runs at a uniform speed of 4 ms$^{-1}$ for 5 minute and then again walks at a speed of 1 ms$^{-1}$ for 3 minutes along a straight track.

Average Speed: 2 ms$^{-1}$, Average Velocity: 2 ms$^{-1}$

Average Speed: 2.5 ms$^{-1}$, Average Velocity: 2.5 ms$^{-1}$

Average Speed: 3 ms$^{-1}$, Average Velocity: 3 ms$^{-1}$

Average Speed: 2 ms$^{-1}$, Average Velocity: 0 ms$^{-1}$

A particle traversed one third the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time and with a velocity $v_2$ for the remaining half of time. Assuming motion to be rectilinear, find the mean velocity of the particle averaged over the whole time of motion from the image given below.

$\frac{v_0 (v_1 + v_2)}{v_0 + v_1 + v_2}$

$\frac{3 v_0 (v_1 + v_2)}{4 v_0 + v_1 + v_2}$

$\frac{2 v_0 (v_1 + v_2)}{2 v_0 + v_1 + v_2}$

$\frac{3 v_0 (v_1 + v_2)}{2 v_0 + 2 v_1 + v_2}$

A body moves along a straight line. Its distance $x$ from a point on its path at a time $t$ after passing that point, is given by $x_t = 8t^2 - 3t^3$ where $x_t$ is in meter and $t$ in second. Find
(a) the instantaneous velocity at $t = 1\text{ s}$
(b) instant and position at which the body is at rest
(c) the acceleration at $t = 4\text{ s}$

(a) $7\text{ ms}^{-1}$; (b) $t = \frac{8}{3}\text{ s}, x = \frac{512}{27}\text{ m}$; (c) $-32\text{ ms}^{-2}$

(a) $7\text{ ms}^{-1}$; (b) $t = \frac{16}{9}\text{ s}, x = \frac{2048}{243}\text{ m}$; (c) $-56\text{ ms}^{-2}$

(a) $5\text{ ms}^{-1}$; (b) $t = \frac{16}{9}\text{ s}, x = \frac{2048}{243}\text{ m}$; (c) $-56\text{ ms}^{-2}$

(a) $7\text{ ms}^{-1}$; (b) $t = 2\text{ s}, x = 8\text{ m}$; (c) $-40\text{ ms}^{-2}$

A body moves along a straight line. Its distance $x$ from a point on its path at a time $t$ after passing that point, is given by $x_t = 8t^2 - 3t^3$ where $x_t$ is in meter and $t$ in second. Find the average velocity and average speed during the interval $t = 0\text{ s}$ to $t = 4\text{ s}$.

Average Velocity: $-16\text{ ms}^{-1}$, Average Speed: $\frac{544}{27}\text{ ms}^{-1}$

Average Velocity: $-16\text{ ms}^{-1}$, Average Speed: $\frac{448}{27}\text{ ms}^{-1}$

Average Velocity: $-8\text{ ms}^{-1}$, Average Speed: $\frac{256}{27}\text{ ms}^{-1}$

Average Velocity: $-16\text{ ms}^{-1}$, Average Speed: $16\text{ ms}^{-1}$

A particle travels along a straight line with a velocity $v = (12 - 3t^2)\text{ ms}^{-1}$, where $t$ is in seconds. When $t = 1\text{ s}$, the particle is located $10\text{ m}$ to the left of the origin. Calculate the
(a) acceleration when $t = 4\text{ s}$
(b) displacement from $t = 0$ to $t = 10\text{ s}$ and
(c) distance the particle travels from $t = 0$ to $t = 10\text{ s}$.

(a) $-24\text{ ms}^{-2}$; (b) $-880\text{ m}$; (c) $912\text{ m}$

(a) $-24\text{ ms}^{-2}$; (b) $-912\text{ m}$; (c) $880\text{ m}$

(a) $-12\text{ ms}^{-2}$; (b) $-880\text{ m}$; (c) $912\text{ m}$

(a) $-24\text{ ms}^{-2}$; (b) $-880\text{ m}$; (c) $880\text{ m}$

A point moving with constant acceleration from $A$ to $B$ in a straight line $AB$ has velocities $u$ and $v$ at $A$ and $B$ respectively.
(a) Find its velocity at $C$, the midpoint of $A$ and $B$.
(b) Find the ratio $\frac{v}{u}$, if time taken from $A$ to $C$ is twice the time to go from $C$ to $B$.

(a) $v_m = \sqrt{\frac{u^2 + v^2}{2}}$, (b) $\frac{v}{u} = 7$

(a) $v_m = \frac{u + v}{2}$, (b) $\frac{v}{u} = 5$

(a) $v_m = \sqrt{u^2 + v^2}$, (b) $\frac{v}{u} = 3$

(a) $v_m = \sqrt{\frac{u^2 + v^2}{2}}$, (b) $\frac{v}{u} = 2$

A body starts with an initial velocity of $10\text{ ms}^{-1}$ and moves along a straight line path with constant acceleration. When the velocity of the body is $50\text{ ms}^{-1}$ the acceleration is reversed in direction. Find the velocity of the particle as it reaches the starting point.

$-70\text{ ms}^{-1}$

$-50\text{ ms}^{-1}$

$-10\text{ ms}^{-1}$

$-30\text{ ms}^{-1}$

Two particles $P$ and $Q$ move in a straight line $AB$ towards each other. $P$ starts from $A$ with a velocity $u_1$ and an acceleration $a_1$. $Q$ starts from $B$ with velocity $u_2$ and an acceleration $a_2$. They pass from each other at midpoint of $AB$ and arrive at other ends of $AB$ with equal velocities. Prove that $(u_1 + u_2)(a_1 - a_2) = 8(a_1 u_2 - a_2 u_1)$.

$(u_1 + u_2)(a_1 - a_2) = 8(a_1 u_2 - a_2 u_1)$

$(u_1 - u_2)(a_1 + a_2) = 8(a_1 u_2 + a_2 u_1)$

$(u_1 + u_2)(a_1 + a_2) = 4(a_1 u_2 - a_2 u_1)$

$(u_1 - u_2)(a_1 - a_2) = 2(a_1 u_2 + a_2 u_1)$

from the image gven below
A man is standing $40\text{ m}$ behind the bus. Bus starts with $1\text{ ms}^{-2}$ constant acceleration and also at the same instant the man starts moving with constant speed $9\text{ ms}^{-1}$. Find the time taken by man to catch the bus.

$8\text{ s}$ and $10\text{ s}$

$4\text{ s}$ and $10\text{ s}$

$5\text{ s}$ and $8\text{ s}$

$6\text{ s}$ and $12\text{ s}$

from the image gven below
A car starts moving rectilinearly, first with an acceleration $a = 5\text{ ms}^{-2}$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $a$, comes to a stop. The total time of motion equals $t = 25\text{ s}$. The average velocity during that time is equal to $v_{av} = 72\text{ kmh}^{-1}$. How long does the car move uniformly?

from the image gven below
A police inspector in a car is chasing a pickpocket an a straight road. The car is going at its maximum speed $v$ (assumed uniform). The pickpocket rides on the motorcycle of a waiting friend when the car is at a distance $d$ away and the motorcycle starts with a constant acceleration $a$. Show that the pick pocket will be caught if $v \ge \sqrt{2ad}$.

$v \ge \sqrt{\frac{ad}{2}}$

A driver takes $0.20\text{ s}$ to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving car at a speed of $54\text{ kmh}^{-1}$ and the brakes cause a deceleration of $6\text{ ms}^{-2}$, find the distance travelled by the car after he sees the need to put the brakes.

The acceleration of a particle as it moves along a straight line is given by $a = (2t - 1) \, ms^{-2}$, where t is in seconds. If $x = 1 \, m$ and $v = 2 \, ms^{-1}$ when $t = 0$, determine the particle's velocity and position when $t = 6 \, s$. Also, determine the total distance the particle travels during this time period, from the image given below.

$v = 32 \, ms^{-1}$, $x = 67 \, m$, $s = 66 \, m$

$v = 30 \, ms^{-1}$, $x = 65 \, m$, $s = 64 \, m$

$v = 32 \, ms^{-1}$, $x = 66 \, m$, $s = 65 \, m$

$v = 34 \, ms^{-1}$, $x = 68 \, m$, $s = 67 \, m$

A particle moves in a straight line with a uniform acceleration $a$. Initial velocity of the particle is zero. Find the average velocity of the particle in first $s$ metre.

$\frac{\sqrt{2as}}{2}$

$\sqrt{\frac{as}{2}}$

In one second a particle goes from point A to point B moving in a semicircle from the image given below. Find the magnitude of average velocity.

$\pi\text{ ms}^{-1}$

$2\pi\text{ ms}^{-1}$

A particle is moving along a straight line such that its position from a fixed point is $x = (12 - 15t^2 + 5t^3)$ m, where t is in seconds. Determine the total distance travelled by the particle from $t = 1$ s to $t = 3$ s. Also, find the average speed of the particle during this time interval, from the image given below.

Total distance = 20 m, average speed = 10 $ms^{-1}$

Total distance = 30 m, average speed = 10 $ms^{-1}$

Total distance = 30 m, average speed = 15 $ms^{-1}$

Total distance = 32 m, average speed = 16 $ms^{-1}$

uestion:
The position $x$ of a particle, in metre, moving along the x-axis depends on the time $t$, in seconds as $x = ct^2 - bt^3$, where $c = 3$ units and $b = 2$ units. Calculate the
(a) units of $c$ and $b$.
(b) time taken by the particle to reach its maximum positive $x$ value.

(a) $c: \text{ms}^{-2}, b: \text{ms}^{-3}$; (b) $1\text{ s}$

(a) $c: \text{ms}^{-1}, b: \text{ms}^{-2}$; (b) $2\text{ s}$

(a) $c: \text{m}, b: \text{m}$; (b) $1\text{ s}$

(a) $c: \text{ms}^{-2}, b: \text{ms}^{-3}$; (b) $3\text{ s}$

The position $x$ of a particle, in metre, moving along the x-axis depends on the time $t$, in seconds as $x = ct^2 - bt^3$, where $c = 3$ units and $b = 2$ units. Calculate the
(c) distance travelled and the displacement of the particle from $t = 0$ to $t = 4\text{ s}$.
(d) velocity and acceleration at $t = 0, 1, 2, 3$ and $4$ second.

(c) Distance: $36\text{ m}$, Displacement: $-80\text{ m}$; (d) $v: 0, 3, 0, -9, -24\text{ ms}^{-1}$; $a: 6, 0, -6, -12, -18\text{ ms}^{-2}$

(c) Distance: $82\text{ m}$, Displacement: $-80\text{ m}$; (d) $v: 0, 0, -12, -36, -72\text{ ms}^{-1}$; $a: 6, -6, -18, -30, -42\text{ ms}^{-2}$

(c) Distance: $82\text{ m}$, Displacement: $-80\text{ m}$; (d) $v: 0, 0, -6, -18, -36\text{ ms}^{-1}$; $a: 6, 0, -6, -12, -18\text{ ms}^{-2}$

(c) Distance: $80\text{ m}$, Displacement: $-80\text{ m}$; (d) $v: 0, 3, 0, -9, -24\text{ ms}^{-1}$; $a: 6, -6, -18, -30, -42\text{ ms}^{-2}$

The position of a particle along a straight line is given by $x = (t^3 - 9t^2 + 15t)\text{ m}$, here $t$ is in second. Determine its maximum acceleration and maximum velocity during the time interval $0 \le t \le 10\text{ s}$.

Maximum Acceleration: $42\text{ ms}^{-2}$, Maximum Velocity: $135\text{ ms}^{-1}$

Maximum Acceleration: $36\text{ ms}^{-2}$, Maximum Velocity: $135\text{ ms}^{-1}$

Maximum Acceleration: $42\text{ ms}^{-2}$, Maximum Velocity: $120\text{ ms}^{-1}$

Maximum Acceleration: $36\text{ ms}^{-2}$, Maximum Velocity: $120\text{ ms}^{-1}$

At time $t = 0$, the position vector of a particle moving in the x-y plane is $5\hat{i}\text{ m}$. At time $t = 0.02\text{ s}$, its position vector has become $5.1\hat{i} + 0.4\hat{j}\text{ m}$. Determine the magnitude of the average velocity ($v_{av}$) during this interval and the angle $\theta$ made by the average velocity with the positive x-axis.

$\vert{}\vec{v}_{av}\vert{} = 20.62\text{ ms}^{-1}$, $\theta = \tan^{-1}(4)$

$\vert{}\vec{v}_{av}\vert{} = 20.62\text{ ms}^{-1}$, $\theta = \tan^{-1}(0.25)$

$\vert{}\vec{v}_{av}\vert{} = 25.00\text{ ms}^{-1}$, $\theta = \tan^{-1}(4)$

$\vert{}\vec{v}_{av}\vert{} = 20.62\text{ ms}^{-1}$, $\theta = \tan^{-1}(2)$

A particle travels along a straight line path such that in 4 s it moves from an initial position $x_A = -8\text{ m}$ to a position $x_B = +3\text{ m}$. Then in another 5 s it moves from $x_B$ to $x_C = -6\text{ m}$. Determine the particle's average velocity and average speed during the 9 s time interval.

Average Velocity: $-0.22\text{ ms}^{-1}$, Average Speed: $2.22\text{ ms}^{-1}$

Average Velocity: $+0.22\text{ ms}^{-1}$, Average Speed: $2.22\text{ ms}^{-1}$

Average Velocity: $+0.22\text{ ms}^{-1}$, Average Speed: $1.89\text{ ms}^{-1}$

Average Velocity: $-0.22\text{ ms}^{-1}$, Average Speed: $1.89\text{ ms}^{-1}$

A particle moves along a horizontal path such that its velocity is given by $v = (3t^2 - 6t) \text{ ms}^{-1}$, where $t$ is the time in seconds. If it is initially located at the origin $O$, determine the
(a) distance travelled by the particle during the time interval $t = 0$ to $t = 3.5\text{ s}$
(b) particle's average velocity and average speed during this time interval.

(a) $14.125\text{ m}$, (b) Average Velocity = $2.625\text{ ms}^{-1}$, Average Speed = $4.036\text{ ms}^{-1}$

(a) $18.375\text{ m}$, (b) Average Velocity = $1.500\text{ ms}^{-1}$, Average Speed = $5.250\text{ ms}^{-1}$

(a) $10.500\text{ m}$, (b) Average Velocity = $3.000\text{ ms}^{-1}$, Average Speed = $3.000\text{ ms}^{-1}$

(a) $12.250\text{ m}$, (b) Average Velocity = $4.200\text{ ms}^{-1}$, Average Speed = $4.200\text{ ms}^{-1}$

The position of a particle along a straight line is given by $x = (1.5t^3 - 13.5t^2 + 22.5t)\text{ m}$, where $t$ is in seconds. Determine the position of the particle when $t = 6\text{ s}$ and the total distance it travels during the $6\text{ s}$ time interval.

Position = $-27\text{ m}$, Total distance = $54\text{ m}$

Position = $-27\text{ m}$, Total distance = $69\text{ m}$

Position = $27\text{ m}$, Total distance = $36\text{ m}$

Position = $27\text{ m}$, Total distance = $69\text{ m}$

The velocity of a particle moving in a straight line decreases at the rate of $3 \text{ ms}^{-1}$ per meter of displacement at an instant when the velocity is $10 \text{ ms}^{-1}$. Calculate the acceleration of the particle at this instant.

$-10 \text{ ms}^{-2}$

$-30 \text{ ms}^{-2}$

$30 \text{ ms}^{-2}$

$-3.33 \text{ ms}^{-2}$