An object moves in a straight line with decel | vedclass

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(B) Given that, deceleration $a = -3 v^{2/3} \,m/s^2$. At initial point $(t=0)$, velocity $u = 8 \,m/s$. We know that acceleration $a = v \frac{dv}{ds

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$4$

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(B) Given that, deceleration $a = -3 v^{2/3} \,m/s^2$. At initial point $(t=0)$, velocity $u = 8 \,m/s$. We know that acceleration $a = v \frac{dv}{ds}$. Substituting the given expression for $a$: $-3 v^{2/3} = v \frac{dv}{ds}$ $-3 v^{2/3} ds = v dv$ $ds = -\frac{1}{3} v^{1 - 2/3} dv = -\frac{1}{3} v^{1/3} dv$. Integrating both sides: $\int_{0}^{s} ds = -\frac{1}{3} \int_{8}^{0} v^{1/3} dv$ $s = -\frac{1}{3} \left[ \frac{v^{4/3}}{4/3} \right]_{8}^{0}$ $s = -\frac{1}{3} \cdot \frac{3}{4} [0 - 8^{4/3}]$ $s = -\frac{1}{4} [0 - (2^3)^{4/3}]$ $s = -\frac{1}{4} [0 - 16] = 4 \,m$. Therefore, the distance travelled by the object before it stops is $4 \,m$.

An object moves in a straight line with deceleration whose magnitude varies with velocity as $3 v^{2/3}$. If at an initial point, the velocity is $8 \,m/s$, then the distance travelled by the object before it stops is (in $\,m$)

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