Physics, asked by Kingironman, 11 months ago

A car accelerates from 36 km/h to 90km/h in5s on a straight rod.what was it's acceleration in m/s and how far did it travel in this time.

Answers

Answered by vishakaa

u=36km/h=36*5/18=10m/sec

u=10m/sec

v=90km/h=90*5/18=25m/sec

v=25m/sec

t=5sec

a=(change in velocity)/t

a=25-5/5

a=4m/sec2

d=ut+1/2at2

d=10*5 + 1/2*5*5*5

d=50+50

=100m

Answered by TheNightHowler

Answer:

Heya mate ✌✌✌

u=36km/h=36*5/18=10m/sec

u=36km/h=36*5/18=10m/secu=10m/sec

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/sec

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec t=5sec

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec t=5sec a=(change in velocity)/t

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec t=5sec a=(change in velocity)/ta=25-10/5

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec t=5sec a=(change in velocity)/ta=25-10/5a=3m/sec2

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec t=5sec a=(change in velocity)/ta=25-10/5a=3m/sec2 d=ut+1/2at2

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec t=5sec a=(change in velocity)/ta=25-10/5a=3m/sec2 d=ut+1/2at2d=10*5 + 1/2*3*5*5

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec t=5sec a=(change in velocity)/ta=25-10/5a=3m/sec2 d=ut+1/2at2d=10*5 + 1/2*3*5*5d=50+37.5

u=36km/h=36*5/18=10m/secu=10m/sec v=90km/h=90*5/18=25m/secv=25m/sec t=5sec a=(change in velocity)/ta=25-10/5a=3m/sec2 d=ut+1/2at2d=10*5 + 1/2*3*5*5d=50+37.5 = 87.5 m

Hope it helps✌✌✌

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