A car accelerates uniformly from 18kmh^-1 to 36 kmh^-1 in 5 sec. ...calculate acceleration and distance covered by car at that time
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solution.
initial velocity, u= 18km/h
final velocity, v= 36km/h
Time taken,t= 5 sec
change time in hour= 5/60 hr or 1/hr
so, Acceleration, a= v-u/t
= 36-18/(1/12)
= 18/(1/12)
= 18* 12
= 216 km/h^2
Now, distance covered during this interval.
s= ut+ 1/2 at^2
= 18*5+ 1/2*216*(5)^2
=> 65+ 108*25
=> 65+ 2700
= 2765 km
______________
hope it helps you☺✌✌
initial velocity, u= 18km/h
final velocity, v= 36km/h
Time taken,t= 5 sec
change time in hour= 5/60 hr or 1/hr
so, Acceleration, a= v-u/t
= 36-18/(1/12)
= 18/(1/12)
= 18* 12
= 216 km/h^2
Now, distance covered during this interval.
s= ut+ 1/2 at^2
= 18*5+ 1/2*216*(5)^2
=> 65+ 108*25
=> 65+ 2700
= 2765 km
______________
hope it helps you☺✌✌
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