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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: one dimensional motion
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[Post New]2 May 2007 00:43:24 IST
    Subject: one dimensional motion
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ishan.maheshwari (161)

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explain the method  to solve the below given question . do not only state the answer :
if the velocity of a car is increased by 20 % , then the minimum distance in which it can be stopped increases by :
   (a) 44 %       (b) 55  %     (c) 66 %      (d) 88 % 
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[Post New]2 May 2007 00:57:24 IST
    Subject: one dimensional motion
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kusum (206)

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hi,
if u increase the veloctity by 20% ur final velocity is 1.2v

since u apply the same force , the acceleration remains the same..therby

V^2=2as
=> v^2/s=constant

1.2^2/s=1/x
s=1.44x
whr x is the initial distace after which the car stops

therefore,
increase%=(1.44x-x)*100/x
=44%
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[Post New]2 May 2007 10:51:10 IST
    Subject: one dimensional motion
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joyfrancis (1504)

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This can also be solved by the work energy theorem.
W = Change in KE

Fnet*X1= w1 = mv^2 / 2
Fnet*X2 = w2 = m(1.2)v^2 / 2
(where w=work done by all forces , v=initial velocity of the car)

Dividing the equations we get
X2 = 36X1/25

So % increase in the dist is 44%
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