A vehicle moves at a speed of 40km/h, it is stopped by applying brakes which produces a uniform acceleration of - 0.6m/s2. How much distance will the vehicle move before coming to stop?
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4
Initial velocity (u)= 40km/h =11.11m/s
Acceleration (a)= -0.6m/s^2
Final velocity (v)= 0 (as it come to rest)
Let time be t
using acceleration formula
a=(v-u)/t
-0.6 = (0-11.11)/t
-0.6t = -11.11
t= 11.11/0.6
=18.51s
Using third equation of motion after arrangement
2as =v^2-u^2
2 (-0.6)(s) = (0)^2 -(11.11)^2
s = 123.43/1.2
=102.86m
So before stopping the body covered 102.86m.
Acceleration (a)= -0.6m/s^2
Final velocity (v)= 0 (as it come to rest)
Let time be t
using acceleration formula
a=(v-u)/t
-0.6 = (0-11.11)/t
-0.6t = -11.11
t= 11.11/0.6
=18.51s
Using third equation of motion after arrangement
2as =v^2-u^2
2 (-0.6)(s) = (0)^2 -(11.11)^2
s = 123.43/1.2
=102.86m
So before stopping the body covered 102.86m.
Answered by
0
Answer:
40
hr
Km
=40×1000/(60×60)
s
m
(=11.1111…).
a=
“delta”t
“delta”v
=
t
(v
f
−v
0
)
=
t
(0−v
0
)
,
t=
a
−v
0
=
−0.6
−11.11
=18.518…sec.,
d=v
0
×
2
t
=
2
11.1111×18.518
=102.9m
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