A particle travels in a straight line so that its velocity, $v$ m/s, is given by \(v = -2t^2 + 6t

Question

A particle travels in a straight line so that its velocity, $v$ m/s, is given by $v = -2t^2 + 6t + 4$, where $t$ seconds is the time after passing $O$. Find an expression in terms of $t$ for

its acceleration,
its displacement from $O$,
the values of $t$ when the speed is $3$ m/s,
the values of $t$ when the displacement of the particle is $0$ m.

Accepted Answer

(a) $a=-4t+6$ (b) $s=-\dfrac23t^3+3t^2+4t$ (c) $t=\dfrac{3+\sqrt{11}}2\text{ s}$ or $t=\dfrac{3+\sqrt{23}}2\text{ s}$ (d) $t=0\text{ s}$ or $t=\dfrac{9+\sqrt{177}}4\text{ s}=5.58\text{ s}$

Kinematics - Differentiating Displacement and Velocity

Kinematics - Differentiating Displacement and Velocity

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