How can you find the following?
$(i)$ Velocity from a displacement-time graph.
$(ii)$ Acceleration from a velocity-time graph.
$(iii)$ Displacement from a velocity-time graph.
$(iv)$ Velocity from an acceleration-time graph.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $(i)$ By finding the slope of the displacement-time graph, where $v = \frac{\Delta \text{displacement}}{\Delta \text{time}}$.
$(ii)$ By finding the slope of the velocity-time graph, where $a = \frac{\Delta \text{velocity}}{\Delta \text{time}}$.
$(iii)$ By finding the area under the velocity-time graph, where $\text{Displacement} = \text{Area under } v-t \text{ graph}$.
$(iv)$ By finding the area under the acceleration-time graph, where $\Delta v = \text{Area under } a-t \text{ graph}$.

Explore More

Similar Questions

The velocity-time graph (Fig.) shows the motion of a cyclist. Find $(i)$ its acceleration $(ii)$ its velocity and $(iii)$ the distance covered by the cyclist in $15\,s$.

What conclusion can you draw from the displacement$-$time graph of a body as shown below?

Study the speed-time graph of a body given below and answer the following questions:
$(i)$ What type of motion is represented by $OA$,$AB$ and $BC$?
(ii) Find positive and negative accelerations of the body.
(iii) Find the distance travelled by the body from $A$ to $B$.

What does the slope of a displacement$-$time graph represent? Can a displacement$-$time graph be parallel to the displacement axis? Give a reason for your answer.

$A$ particle accelerates from rest at a constant rate for some time and attains a constant velocity of $8\, m s^{-1}$. Afterwards,it decelerates at a constant rate and comes to rest. If the total time taken is $4\, s$,the distance travelled is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo