A bullet is shot horizontally and its distance S cm at time t second is given by $\mathrm{S}=1200 \mathrm{t}-15 \cdot \mathrm{t}^2$, then the distance covered by the bullet when it comes to the rest, is

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Aicharya · Accepted Answer

$s=1200 t-15 t^2$ $\operatorname{Velocity}(\mathrm{v})=\frac{\mathrm{ds}}{\mathrm{dt}}=1200-30 \mathrm{t}$ When bullet stopped, $\frac{\mathrm{ds}}{\mathrm{dt}}=0$ $\Rightarrow t=40$
Hence, required distance $=1200(40)-15 \times(40)^2=24000 \mathrm{~cm}$

A bullet is shot horizontally and its distance S cm at time t second is given by $\mathrm{S}=1200…

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