A truck is moving at a speed of 72 kmh-1 on a straight road. The driver can produce deceleration of 2 ms-2 by applying brakes. The stopping distance of truck is 13x m, if the reaction time of the driver is 0.2 s. The value of x is_________

Solution:

$\begin{array}{l} ∵ v^{2} = u^{2} + 2 a s \\ \Rightarrow 0^{2} = {(72 \times \frac{5}{18})}^{2} - 2 \times 2 s_{0} \\ ∴ s_{0} = \frac{400}{4} = 100 m \end{array}$

$\begin{array}{l} Thus, stopping distance = (72 \times \frac{5}{18}) \times 0.2 + s_{0} \\ \begin{array}{ll} = 4 + 100 = 104 = 13 x \\ ∴ x = \frac{104}{13} = 8 \end{array} \end{array}$

A truck is moving at a speed of 72 kmh^-1 on a straight road. The driver can produce deceleration of 2 ms^-2 by applying brakes. The stopping distance of truck is 13x m, if the reaction time of the driver is 0.2 s. The value of x is_________

Solution:

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