A car starts from rest and acquires a velocity of 54mkmh-1 in 2 minutes find acceleration and distance travelled by car during this time assume,motion of car is uniform
Answers
Answered by
2
Given,
(u) initial velocity=0 m/sec
(v) final velocity=54 mkm/h
= ( 54 m×1000 m )/(60×60)sec
=(54000m^2)/3600 sec
=15m^2/sec
(t) time=120 sec
now,
1 . (a) acceleration=(v-u)/t
=(15 m^2 -0)/120 m/s^2
=(1500 m-0)/120 m/s^2
=(12.5) m/s^2
2. i have used second equation of motion for calculating distance (s)
= (s) = ut+ at^2
=(s)=(0×120) +(12.5 ×(120)^2(m)
=(s)=0+180,000 m
=(s) = 180 km
clearly,acceleration is 12.5 m and distance is 180,000 m i.e 180 km.
(u) initial velocity=0 m/sec
(v) final velocity=54 mkm/h
= ( 54 m×1000 m )/(60×60)sec
=(54000m^2)/3600 sec
=15m^2/sec
(t) time=120 sec
now,
1 . (a) acceleration=(v-u)/t
=(15 m^2 -0)/120 m/s^2
=(1500 m-0)/120 m/s^2
=(12.5) m/s^2
2. i have used second equation of motion for calculating distance (s)
= (s) = ut+ at^2
=(s)=(0×120) +(12.5 ×(120)^2(m)
=(s)=0+180,000 m
=(s) = 180 km
clearly,acceleration is 12.5 m and distance is 180,000 m i.e 180 km.
Similar questions