7. A car acquires a velocity of 72km/h in 10 seconds starting from rest. Find (i) acceleration (ii) the average velocity, and (iii) the distance travelled in tim
Answers
Answered by
1
Given,
initial velocity(u)=0 m/s
final velocity(v)=72 km/h i.e 72×5/18 m/sec
i.e 20 m/s
time(t)=10
now,
(i) the acceleration (a)=(v-u)/t
(a)=(20-0)/10 m/s^2
(a)=2 m/s^2
(ii)Average velocity= (v+u)/2
=(20+0)/2 m/s
=10 m/s
(iii) i have applied second equation of motion for find distance(s)
=(s)=ut+1/2 at^2
=(s)=(0×10) +(1/2×2×10^2 m
=(s)=100 m
clearly,acceleration is 2 m/s^2 ,average velocity is 10m/s and distance is 100m.
initial velocity(u)=0 m/s
final velocity(v)=72 km/h i.e 72×5/18 m/sec
i.e 20 m/s
time(t)=10
now,
(i) the acceleration (a)=(v-u)/t
(a)=(20-0)/10 m/s^2
(a)=2 m/s^2
(ii)Average velocity= (v+u)/2
=(20+0)/2 m/s
=10 m/s
(iii) i have applied second equation of motion for find distance(s)
=(s)=ut+1/2 at^2
=(s)=(0×10) +(1/2×2×10^2 m
=(s)=100 m
clearly,acceleration is 2 m/s^2 ,average velocity is 10m/s and distance is 100m.
Similar questions