(A) The slope of the distance-time $(s-t)$ graph represents the velocity of the particle.Initial velocity at time $t$ is $v_i = \tan(45^{\circ}) = 1 \

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(A) The slope of the distance-time $(s-t)$ graph represents the velocity of the particle.Initial velocity at time $t$ is $v_i = \tan(45^{\circ}) = 1 \, m/s$.Final velocity after one second is $v_f = \tan(60^{\circ}) = \sqrt{3} \, m/s$.The average acceleration is defined as the change in velocity divided by the time interval:$a_{av} = \frac{v_f - v_i}{\Delta t}$.Substituting the values:$a_{av} = \frac{\sqrt{3} - 1}{1} = (\sqrt{3} - 1) \, m/s^2$.

Class 11 Physics Motion in a Straight Line Acceleration and its graph

Medium

The distance-time graph of a particle at time $t$ makes an angle of $45^{\circ}$ with the time axis. After one second,it makes an angle of $60^{\circ}$ with the time axis. What is the average acceleration of the particle?

A
$\sqrt{3}-1$
B
$\sqrt{3}+1$
C
$\sqrt{3}$
D
$1$

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Motion in a Straight Line

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Acceleration and its graph

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The distance-time graph of a particle at time $t$ makes an angle of $45^{\circ}$ with the time axis. After one second,it makes an angle of $60^{\circ}$ with the time axis. What is the average acceleration of the particle?

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The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k} \text{ m/s}$. The magnitude of acceleration at point $(1, 2, 4)$ is . . . . . . $\text{m/s}^2$.

If velocity and acceleration are in opposite directions in one-dimensional motion,what happens to the magnitude of velocity?

The acceleration at the end of $2 \ s$ of a particle whose motion is represented by the equation $S = 4t^3 - 8t^2 + 5t + 4$ is $......$ (in $m \ s^{-2}$)

An object with a mass $10 \, kg$ moves at a constant velocity of $10 \, m/s$. $A$ constant force then acts for $4 \, s$ on the object and gives it a speed of $2 \, m/s$ in the opposite direction. The acceleration produced in it is ........ $m/s^2$.

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