A car accelerates uniformly from 18km/hr to 36 km/hr in 5 sec.calculate the distance covered by the car in 5 sec.
Answers
Answered by
101
Acceleration = Change in Velocity/Time
Change in Velocity = 36-18 = 18 km/h=5 m/s
Time= 5 Seconds
Acceleration = 5/5= 1 m/s^2
Equation of motion,s=ut+(1/2)at^2
u=18 km/h=5 m/s
t=5 s
a=1 m/s^2
s= (5*5)+(1/2*1*5*5)
s=25+12.5
i.e.,
s=37.5 m
Hope you are clear with my explanations
Change in Velocity = 36-18 = 18 km/h=5 m/s
Time= 5 Seconds
Acceleration = 5/5= 1 m/s^2
Equation of motion,s=ut+(1/2)at^2
u=18 km/h=5 m/s
t=5 s
a=1 m/s^2
s= (5*5)+(1/2*1*5*5)
s=25+12.5
i.e.,
s=37.5 m
Hope you are clear with my explanations
Answered by
70
u= initial velocity= 18 km/h = 18 x 5/18= 5 m/s
v= final velocity = 36 km/h = 36 x 5/18 = 10 m/s
t= time = 5 seconds
a = acceleration = (v-u)/t = (10-5)/5 = 5/5 = 1 m/s²
s= distance = ?
using second equation of motion,
s= ut+ 1/2 at²
we get,
s= 5 x 5 + 1/2 x 1 x 5²
s = 25 + 12.5
s= 37.5 m
∴ the distance covered is 37.5m
v= final velocity = 36 km/h = 36 x 5/18 = 10 m/s
t= time = 5 seconds
a = acceleration = (v-u)/t = (10-5)/5 = 5/5 = 1 m/s²
s= distance = ?
using second equation of motion,
s= ut+ 1/2 at²
we get,
s= 5 x 5 + 1/2 x 1 x 5²
s = 25 + 12.5
s= 37.5 m
∴ the distance covered is 37.5m
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