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A car accelerates uniformly from 36 km/h to 45 km/h in 15 s. Calculate the distancecovered by the car in that time. |
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Answer»
Given parameters Time taken (t) = 5 sec Initial velocity (u) =18 km/hour
\(u = \frac{18\times 1000}{60\times 60S}\)
u = 5 m/s Final velocity (v) =36km/hour
\(v = \frac{36\times 1000}{60\times 60s}\)We know that distance travelled is CALCULATED by the formula Distance travelled S = u t + (1/2) a × t2 Substituting the given values in the above equation we get, S = 5 × 5 + (1/2) × 1 × 52 S = 25 + (1/2) × 25 S = 25 + 12.5 S = 37.5 m Hence acceleration a =1 m/s2 Distance travelled S =37.5 m Pls MARK as brainlest...
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