A car starting from rest accelerates at the rate f through a distance s,then continues at constant speed for time t and accelerates at the rate f/2 to come to rest if the total distance travelled is 15s then s=?
Ans:S=ft^2/72
how it came plz ans with simple method
Answers
Answer:
S=(ft^2)/2
V.=squere root of(2Sf)
During retardation S2=2S
During constant velocity 15S -3S = 12S = vo
. '' .S= (ft^2)/72
Answer:
Correct option is
A
s=
72
ft
2
If t
1
be the time for which the car accelerates at the rate f from rest, the distance traveled in time t
1
is
s=s
1
=0(t
1
)+
2
1
ft
1
2
=
2
1
ft
1
2
....(1) (using formula
s=ut+
2
1
at
2
)
and the velocity at time t
1
is v
1
=ft
1
So, the distance traveled in time t will be s
2
=v
1
t=ft
1
t ...(2)
If t
2
be the time for which the car decelerates at the rate f/2 to come rest, the distance traveled in time t
2
is given by,
0
2
−v
1
2
=2(−f/2)s
3
(using formula v
2
−u
2
=2as)
or s
3
=(ft
1
)
2
/f=ft
1
2
...(3)
using (1), s
3
=2s
1
=2s
Given, s
1
+s
2
+s
3
=15s
or s+ft
1
t+2s=15s
or ftt
1
=12s
or 12s=ftt
1
..(4)
(4)/(1)⇒
s
12s
=
(1/2)ft
1
2
ftt
1
or t
1
=t/6
From (1), s=(1/2)f(t/6)
2
=ft
2
/72
Explanation:
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