A car has initial velocity of 72 km/h.it is accelerated at 2m/s.calculate the final velocityand the distance covered after 3 sec
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initial velocity u = 72 km/h = 72×5/18 =4×5=20m/s
acceleration a = 2m/s^2
time t = 3 sec
final velocity v =?
distance S =?
from first equation of motion,
v = u + a.t
v = 20 + 2 × 3
v = 20+ 6
v = 26 m/s
from second equation of motion,
S = u.t + 1/2 a.t^2
S = 20×3 +1/2×2×3×3
S = 60 + 9
S = 69 m
acceleration a = 2m/s^2
time t = 3 sec
final velocity v =?
distance S =?
from first equation of motion,
v = u + a.t
v = 20 + 2 × 3
v = 20+ 6
v = 26 m/s
from second equation of motion,
S = u.t + 1/2 a.t^2
S = 20×3 +1/2×2×3×3
S = 60 + 9
S = 69 m
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