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Standard XII

Physics

Displacement in 2D Motion

The velocity ...

Question

The velocity of a particle moving along a line is $v = t^{2} - 5 t + 6$ , from $t = 4$ to $t = 8$ . What is the relation between distance $(S)$ and displacement $(Δ X)$

S > | Δ X |

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S < | Δ X |

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S = | Δ X |

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S > | Δ X |^{2}

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Solution

The correct option is C $S = | Δ X |$
Plot $v$ - t graph, after $t = 3$ , velocity $v$ is positive. Area under $v - t$ graph is also positive.

So distance and displacement are equal.

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Similar questions

Velocity-time equation of a particle moving in a straight line is v = 2t - 4 for t ≤ 2 s and v = 4-2t for t> 2 s. The distance travelled by the particle in the time interval from t = 0 to t = 4 s is (Here, t is in second and v in m/s)

Q. A particle moving along a straight line with constant acceleration of

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Q. A particle moving along a straight line with constant acceleration of

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Q. The displacement of a particle moving in a straight line is described by the relation,

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Figure shows the velocity-time graph of a particle moving in a straight line.

(i) State the nature of motion of particle.

(ii) Find the displacement of particle at t=6 s.

(iii) Does the particle change its direction of motion?

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The velocity of a particle moving along a line is v=t2−5t+6, from t=4 to t=8. What is the relation between distance (S) and displacement (ΔX)

The correct option is C S=|ΔX|Plot v - t graph, after t=3, velocity v is positive. Area under v−t graph is also positive. So distance and displacement are equal.

The velocity of a particle moving along a line is $v = t^{2} - 5 t + 6$ , from $t = 4$ to $t = 8$ . What is the relation between distance $(S)$ and displacement $(Δ X)$

The correct option is C $S = | Δ X |$
Plot $v$ - t graph, after $t = 3$ , velocity $v$ is positive. Area under $v - t$ graph is also positive.

So distance and displacement are equal.