A car is driven for 0.8 s. If the car travelling initially with 36 km − h⁻¹ is stopped by the driver after observing a signal by deceleration of 5 m − s−², then total distance travelled by the car before coming to rest, is
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so here t=0.8 sec
u=36km/hr which has to be converted in m/s
so multiply 5/18 with 36 to get 10 m/s as the initial velocity.
a= -5m/s (here i have taken negative sign as the body is retarding)
v=0 (this information is more than enough)
now apply the 2nd or 3rd equation of motion(whatever you want)
i am using 2nd equation of motion.
s=ut+1/2at^2
s=(10*0.8)+(1/2*-5*0.8*0.8)
at the end s will come out to be 9.6 m
THANK YOU:)
u=36km/hr which has to be converted in m/s
so multiply 5/18 with 36 to get 10 m/s as the initial velocity.
a= -5m/s (here i have taken negative sign as the body is retarding)
v=0 (this information is more than enough)
now apply the 2nd or 3rd equation of motion(whatever you want)
i am using 2nd equation of motion.
s=ut+1/2at^2
s=(10*0.8)+(1/2*-5*0.8*0.8)
at the end s will come out to be 9.6 m
THANK YOU:)
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