A truck running at 90 text{ km h}^{-1} is bro | vedclass

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(A) Given: Initial velocity $u = 90 	ext{ km h}^{-1} = 90 	imes \frac{5}{18} 	ext{ m s}^{-1} = 25 	ext{ m s}^{-1}$.Final velocity $v = 0 	ext{ m

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Retardation = $12.5 	ext{ m s}^{-2}$,Time = $2 	ext{ s}$

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(A) Given: Initial velocity $u = 90 	ext{ km h}^{-1} = 90 	imes \frac{5}{18} 	ext{ m s}^{-1} = 25 	ext{ m s}^{-1}$.Final velocity $v = 0 	ext{ m s}^{-1}$ (as the truck comes to rest).Distance $S = 25 	ext{ m}$.$(i)$ To find retardation $(a)$: Using the third equation of motion,$v^{2} - u^{2} = 2aS$.$(0)^{2} - (25)^{2} = 2 	imes a 	imes 25$.$-625 = 50a$.$a = -\frac{625}{50} = -12.5 	ext{ m s}^{-2}$.Retardation is the magnitude of negative acceleration,so retardation = $12.5 	ext{ m s}^{-2}$.(ii) To find time $(t)$: Using the first equation of motion,$v = u + at$.$0 = 25 + (-12.5) 	imes t$.$12.5t = 25$.$t = \frac{25}{12.5} = 2 	ext{ s}$.

$A$ truck running at $90 \text{ km h}^{-1}$ is brought to rest over a distance of $25 \text{ m}$. Calculate the retardation and time for which brakes are applied.

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