A car acquires a velocity of 72km/h in 10sec starting fromrest. Find (a)the acceleration (b) Average velocity (c) the distance travelled in this time
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Given,
initial velocity(u)=0m/s
final velocity( v)=72 km/h i.e( 72×5/18) m/s
time (t) =10 sec
now,
(a) the acceleration (a)= (v-u)/ t
(a) =( 20-0)/10 m/s^2
(a)=2m/s^2
(b) Average velocity=(v+u)/2
=(20+0) /2 m/s
=10 m/s
(c) I have use second equation of motion for calculating distance(s) in given case,
= (s)= (0×10) + (1/2×2 ×10^2) m
=(s) =0 + 100 m
=(s) = 100 m
clearly,the acceleration is 2 m/s^2 ,Average velocity is 10 m/s and distance is 100m
initial velocity(u)=0m/s
final velocity( v)=72 km/h i.e( 72×5/18) m/s
time (t) =10 sec
now,
(a) the acceleration (a)= (v-u)/ t
(a) =( 20-0)/10 m/s^2
(a)=2m/s^2
(b) Average velocity=(v+u)/2
=(20+0) /2 m/s
=10 m/s
(c) I have use second equation of motion for calculating distance(s) in given case,
= (s)= (0×10) + (1/2×2 ×10^2) m
=(s) =0 + 100 m
=(s) = 100 m
clearly,the acceleration is 2 m/s^2 ,Average velocity is 10 m/s and distance is 100m
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