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Standard XII

Physics

Impulsive Forces in Collision

The velocity ...

Question

The velocity $-$ time graph of a linear motion is shown in figure. The distance from the starting point after $8$ sec is

16 m

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19 m

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8 m

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5 m

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Solution

The correct option is D $5 m$
Displacement $=$ Area under $v - t$ graph

$= \frac{1}{2} (4 + 2) \times 4 - \frac{1}{2} (4 + 3) \times 2$

$= 12 - 7 = 5 m$

Final answer : (a)

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Impulsive Forces in Collision

Standard XII Physics

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The velocity − time graph of a linear motion is shown in figure. The distance from the starting point after 8 sec is

The correct option is D 5 mDisplacement = Area under v−t graph =12(4+2)×4−12(4+3)×2 =12−7=5 m Final answer : (a)

The velocity $-$ time graph of a linear motion is shown in figure. The distance from the starting point after $8$ sec is

The correct option is D $5 m$
Displacement $=$ Area under $v - t$ graph

$= \frac{1}{2} (4 + 2) \times 4 - \frac{1}{2} (4 + 3) \times 2$

$= 12 - 7 = 5 m$

Final answer : (a)