Acceleration-time graph of a particle moving on a straight line is given below. The units of the axes are in SI system. If initial velocity of particle is 2 m/s in the direction of acceleration, then its velocity at end of 4 sec is
A
8 m/s
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B
12 m/s
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C
10 m/s
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D
14 m/s
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Solution
The correct option is D 14 m/s Area under a-t graph gives change in velocity.
So, ΔV=Area=4×2+2×2=8+4=12m/s ΔV=Vf−Vi or 12=Vf−2⇒Vf=14m/s