Question

Х The velocity-time graph of a train traveling between two stations is plotted below:
Find th e train's acceleration (in ms^-2)in each of the time intervals OX,XY,YZ respectively
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Answer 1

According to the velocity-time graph we are asked to find out the acceleration in each section

Explanation: In a velocity time graph, the slope of graph tell us about the acceleration. If the slope is high then the acceleration is positive, if the solve is low then the acceleration is negative and if the slope is parallel to time axis then there is no acceleration taking place!

Required solution:

Finding acceleration in section OX!

Here: the final velocity is 12 m/s, initial velocity is 0 m/s and time taken is 6 seconds (∵ 6-0)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{12-0}{6} \\ \\ :\implies \sf a \: = \dfrac{12}{6} \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2}$

Here, acceleration = 2 m/s sq.

Finding acceleration in section XY!

Here: Final velocity is 12 m/s and initial velocity is 12 m/s and the the time taken is 12 seconds (∵ 18-6)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{12-12}{12} \\ \\ :\implies \sf a \: = \dfrac{0}{12} \\ \\ :\implies \sf a \: = 0 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 0 \: ms^{-2}$

Explanation: Surprised that acceleration is zero m/s sq. here? Actually section XY is not telling us about actual acceleration. As acceleration is defined as the change in velocity per time but in this section the final and initial velocity is same. And we also already know that is velocity is constant then acceleration is 0!

Here, acceleration = 0 m/s sq.

Acceleration in section YZ!

Here: Final velocity is 0 m/s, initial velocity is 12 m/s and time is 3 second (∵ 21-18)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-12}{3} \\ \\ :\implies \sf a \: = \dfrac{-12}{3} \\ \\ :\implies \sf a \: = -4 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -4 \: ms^{-2}$

Here, acceleration = -4 m/s sq.