Answer:
It is
better to apply brake than to turn sharply for the reason discussed below.
Let m be the mass of the car. When the driver applies the
brakes, let the car stops at distance x. Then retardation, \[\text{a}={{\text{v}}^{\text{2}}}/\text{2x}\]
Retarding force, =\[\text{F
}=\text{ma}=\frac{m{{v}^{2}}}{2x}\] or\[\text{x }=\frac{m{{v}^{2}}}{2F}\].
There will be no collision if \[\text{x}<\text{r}\] or \[\frac{m{{v}^{2}}}{2F}~\le
\text{r}\] or \[\text{F}\ge \frac{m{{v}^{2}}}{2r}\] ....(i)
If the driver takes a sharp turn of radius x, then
centripetal force on car is,
\[\text{F }\!\!'\!\!\text{ =}\frac{m{{v}^{2}}}{x}\]or \[\text{x}\,\text{=}\frac{m{{v}^{2}}}{F'}\]
To avoid collision, \[\text{x }\le \text{r}\] so \[\frac{m{{v}^{2}}}{F'}~\le
\text{r}\] or F' ³ \[\text{F }\!\!'\!\!\text{ }\ge \frac{m{{v}^{2}}}{r}\] ....(ii)
From (i) and (ii) we note that \[\text{F }=\text{
}\frac{\text{F}'}{2}\].
It means to avoid collision braking force required is half
the centripetal force. Therefore, braking is better.
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