A car moving at 72km/h is stopped in 5 seconds.If it retards uniformly find the distance travelled by the car before it stops.Also find its retardation.
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Answers
Step-by-step explanation:
▪️Acceleration = (final velocity - initial velocity) / time
▪️Initial velocity(Vi) = 72km/h = 72,000 m/h. (A kilometre is 1000 metres)
▪️There are 3600 seconds in an hour, so
▪️Vi = 72,000/3600= 20m/sec. Final velocity(Vf) = 0
▪️Vf - Vi = 0–20 = -20m/sec.
▪️So deceleration is -20/10 = -2m/sec/sec (metres per second squared)
▪️(The minus indicates deceleration)
▪️What this means is that the car loses speed at a uniform rate of 2m/sec,
▪️so after 1 second, velocity = 18m/sec, after 2 secs, 16m/sec…..etc.
Hopes it help you❤️❤️
Step-by-step explanation:
u = 72 km/h = 20 m/s
t = 5 s
As car is retarding , the final velocity v = 0 m/s
a = (v-u)/t
= (0-20)/5
= -20/5
= -4 m/s^2
retardation r = -a
r = 4 m/s^2
We know ,
2as = v^2 - u^2
2 x -4 x s = 0 - (20)^2
-8 x s = -400
s = 50 m
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