A car travelling at 108 $k m h^{- 1}$ has its speed reduced to 36 $k m h^{- 1}$ after travelling a distance of 200 m. Find the retardation (assumed uniform) and time taken for this process.

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Solution

$V_{2} = 108 \times \frac{5}{18} = 30 m / s$ $V_{1} = 36 \times \frac{5}{18} = 10 m / s$
$\Rightarrow V^{2} = U^{2} + 2 a s$
$\Rightarrow (30)^{2} = (30)^{2} + 2 \times a \times 200$
$\Rightarrow a = - 2 m / s$

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A car travelling at 108 kmh−1 has its speed reduced to 36 kmh−1 after travelling a distance of 200 m. Find the retardation (assumed uniform) and time taken for this process.

V2=108×518=30m/s V1=36×518=10m/s⇒V2=U2+2as⇒(30)2=(30)2+2×a×200⇒a=−2m/s

A car travelling at 108 $k m h^{- 1}$ has its speed reduced to 36 $k m h^{- 1}$ after travelling a distance of 200 m. Find the retardation (assumed uniform) and time taken for this process.

$V_{2} = 108 \times \frac{5}{18} = 30 m / s$ $V_{1} = 36 \times \frac{5}{18} = 10 m / s$
$\Rightarrow V^{2} = U^{2} + 2 a s$
$\Rightarrow (30)^{2} = (30)^{2} + 2 \times a \times 200$
$\Rightarrow a = - 2 m / s$