A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate 
to come to rest. If the total distance travelled is 15 S, then:
ft 2
ft 2
ft 2
Text Solution
Verified by ExpertsC
v = ft 1
and the final velocity of OA = initial velocity of BC
ft 1 = 
t 2
∴ t 2 = 2t 1
In graph
S 1 = 
ft 12 ..... (i)
Given, S 1 = S
S 2 = (ft 1 )t
S 3 = 
. 
(2t 1 ) 2
Thus S 1 +S 2 +S 3 = 15 S
S + (ft 1 )t + 2S = 15S 
(tf 1 ) t = 125 ..... (ii)
From eq. (i) and (ii) and (iii), we have
∴ t 1 = 
From Eq. (i), we get
∴ S = 
f(t 1 ) 2
∴ S = 
f(t/6) 2 = 
ft 2
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