Question

a car starting from rest at the rate of f through a distance S then continues tomove at a constant speed for time t and then decelerates at the rate f/2 to come to test. 1if the total distance is 5 S,then​

Answer 1

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