An automobile,travelling at 40, km/h,can be s | vedclass

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(B) Using the equation of motion $v^2 - u^2 = 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration (deceleration),

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$160$

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(B) Using the equation of motion $v^2 - u^2 = 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration (deceleration),and $s$ is the stopping distance.For the first case: $u_1 = 40\, km/h$,$v_1 = 0$,$s_1 = 40\, m$.$0^2 - u_1^2 = 2a(40) \implies a = -u_1^2 / 80$.Since the deceleration $a$ is constant for the same braking force,we have $s \propto u^2$.Therefore,$s_2 / s_1 = (u_2 / u_1)^2$.Given $u_2 = 80\, km/h$,$u_1 = 40\, km/h$,and $s_1 = 40\, m$.$s_2 = 40 	imes (80 / 40)^2 = 40 	imes 2^2 = 40 	imes 4 = 160\, m$.The minimum stopping distance is $160\, m$.

An automobile,travelling at $40\, km/h$,can be stopped at a distance of $40\, m$ by applying brakes. If the same automobile is travelling at $80\, km/h$,the minimum stopping distance,in metres,is (assume no skidding)..........$m$

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