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

8 m/s

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12 m/s

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10 m/s

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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, $Δ V = A r e a = 4 \times 2 + 2 \times 2 = 8 + 4 = 12 m / s$
$Δ V = V_{f} - V_{i}$ or $12 = V_{f} - 2 \Rightarrow V_{f} = 14 m / s$

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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

The correct option is D 14 m/sArea under a-t graph gives change in velocity. So, ΔV=Area=4×2+2×2=8+4=12 m/s ΔV=Vf−Vi or 12=Vf−2⇒Vf=14 m/s

The correct option is D 14 m/s
Area under a-t graph gives change in velocity.
So, $Δ V = A r e a = 4 \times 2 + 2 \times 2 = 8 + 4 = 12 m / s$
$Δ V = V_{f} - V_{i}$ or $12 = V_{f} - 2 \Rightarrow V_{f} = 14 m / s$