The velocity-time graph of a train travelling between two stations, is plotted as below:

Find train's average acceleration ( in $m / s^{2}$ ) in each of the time intervals OX, XY and YZ respectively.

-2, 1 , 4

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2, 1, -4

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-2, 0 , 4

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2 , 0 , -4

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Solution

The correct option is D
2 , 0 , -4

Average acceleration = $\frac{F i n a l V e l o c i t y - I n i t i a l V e l o c i t y}{T i m e}$

Acceleration in OX $= \frac{12 - 0}{6}$ = 2 $\frac{m}{s^{2}}$

Acceleration in XY $= \frac{12 - 12}{12}$ = 0 $\frac{m}{s^{2}}$

Acceleration in YZ $= \frac{0 - 12}{3}$ = -4 $\frac{m}{s^{2}}$

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The velocity-time graph of a train travelling between two stations, is plotted as below: Find train's average acceleration ( in m/s2 ) in each of the time intervals OX, XY and YZ respectively.

The correct option is D 2 , 0 , -4 Average acceleration = Final Velocity−Initial VelocityTime Acceleration in OX=12−06 = 2 ms2 Acceleration in XY =12−1212 = 0 ms2 Acceleration in YZ =0−123= -4 ms2

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