The speed of a car is reduced from 90 km/hr to 36km/hr in 5s what is the distance travelled by the car during this time
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initial velocity u=90kmph=25m/sec.
final velocity=36kmph=10m/sec.
we Know that ......
v=u+at.
so 10=25+5a
a=-3m/s^2.
and distance =ut +1/2at^2
=25(5)+1/2(-3)(5^2)
=125-75/2
=175/2
=87.5m
final velocity=36kmph=10m/sec.
we Know that ......
v=u+at.
so 10=25+5a
a=-3m/s^2.
and distance =ut +1/2at^2
=25(5)+1/2(-3)(5^2)
=125-75/2
=175/2
=87.5m
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