A car takes 8 s to go from v = 0 m/s to v = 20 m/s atapproximately constant acce

Question

A car takes 8 s to go from v = 0 m/s to v = 20 m/s atapproximately constant acceleration. Calculate: (i) The acceleration of the car a = v / t = 20 m/s / 8 s = 2.5m/s2 (ii) The distance travelled by the car d = 1/2 (at2) = 1/2 (2.5 x 64) = 80 m As I am preparing for my end of year exams, could someoneplease tell me if I have calculated the above correctly or not. Anyadvice/suggestions would be most helpful. Many thanks in advance. Cheers Tony A car takes 8 s to go from v = 0 m/s to v = 20 m/s atapproximately constant acceleration. Calculate: (i) The acceleration of the car a = v / t = 20 m/s / 8 s = 2.5m/s2 (ii) The distance travelled by the car d = 1/2 (at2) = 1/2 (2.5 x 64) = 80 m As I am preparing for my end of year exams, could someoneplease tell me if I have calculated the above correctly or not. Anyadvice/suggestions would be most helpful. Many thanks in advance. Cheers Tony

Explanation / Answer

           Giventhat the initial velocity of car is u = 0 m/s            Finalvelocity of car is v = 20 m/s            Timetaken is t = 8s -----------------------------------------------------------------------       The acceleration is a = ( v -u ) / t          (or) v / t                                       = 20m/s / 8s                                       = 2.5m/s2    The distance traveled by the car in timet is from equation of motion     S = ut + (1/2)at2      = 0 + (1/2)at2   ( sinceu = 0 m/s )    = 80 m

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