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Standard XII

Physics

v-t Graph

The velocity ...

Question

The velocity time graph of a particle moving along a straight line has the form of a parabola $t^{2} - 6 t + 8$ m/s. Find distance travelled (in m) between $t = 0$ and $t = 3$ seconds :

\frac{22}{3}

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\frac{22}{4}

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22

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6

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Solution

The correct option is D $6$
Here, velocity $v = t^{2} - 6 t + 8$
So, $\frac{d x}{d t} = t^{2} - 6 t + 8$
Integrating, $x = \int_{0}^{3} (t^{2} - 6 t + 8) d t = [t^{3} / 3 - 3 t^{2} + 8 t]_{0}^{3} = 9 - 27 + 24 = 6 m$

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Similar questions

Q. The velocity time graph of a particle moving along a straight line has the form of a parabola

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The velocity time graph of a particle moving along a straight line has the form of a parabola t2−6t+8 m/s. Find distance travelled (in m) between t=0 and t=3 seconds :

The correct option is D 6Here, velocity v=t2−6t+8So, dxdt=t2−6t+8Integrating, x=∫30(t2−6t+8)dt=[t3/3−3t2+8t]30=9−27+24=6m

The velocity time graph of a particle moving along a straight line has the form of a parabola $t^{2} - 6 t + 8$ m/s. Find distance travelled (in m) between $t = 0$ and $t = 3$ seconds :

The correct option is D $6$
Here, velocity $v = t^{2} - 6 t + 8$
So, $\frac{d x}{d t} = t^{2} - 6 t + 8$
Integrating, $x = \int_{0}^{3} (t^{2} - 6 t + 8) d t = [t^{3} / 3 - 3 t^{2} + 8 t]_{0}^{3} = 9 - 27 + 24 = 6 m$