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(A) The speed of a particle is minimum when the component of acceleration along the velocity vector is zero.Let the initial velocity be $\vec{v}_0$ an

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(A) The speed of a particle is minimum when the component of acceleration along the velocity vector is zero.Let the initial velocity be $\vec{v}_0$ and acceleration be $\vec{a}$.The angle between $\vec{v}_0$ and $\vec{a}$ is $\phi = 143^o$.The component of acceleration along the direction of initial velocity is $a_{\parallel} = a \cos(143^o)$.Since $\cos(143^o) = \cos(180^o - 37^o) = -\cos(37^o) = -0.8$,we have $a_{\parallel} = 4 	imes (-0.8) = -3.2\, m/s^2$.The speed $v(t)$ at time $t$ is given by $v(t) = v_0 + a_{\parallel} t$.For minimum speed,the tangential acceleration must be zero,but here the acceleration is constant. The speed is minimum when the velocity vector is perpendicular to the acceleration vector.Let $\vec{v}(t) = \vec{v}_0 + \vec{a}t$.For minimum speed,$\vec{v}(t) \cdot \vec{a} = 0$.$(\vec{v}_0 + \vec{a}t) \cdot \vec{a} = 0$$\vec{v}_0 \cdot \vec{a} + a^2 t = 0$$v_0 a \cos(143^o) + a^2 t = 0$$t = -\frac{v_0 \cos(143^o)}{a} = -\frac{40 	imes (-0.8)}{4} = \frac{32}{4} = 8\, s$.

The figure shows the velocity and the acceleration of a point-like body at the initial moment of its motion. The direction and the absolute value of the acceleration remain constant. Find the time when the speed becomes minimum. (Given: $a = 4\, m/s^2$,$v_0 = 40\, m/s$,$\phi = 143^o$)

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