Posted by **fahd** on Friday, June 8, 2012 at 1:23pm.

A CAR ACCELERATES TO 36M/S FOR 4 SECS BEFORE DECELERATING TO 32M/S FOR THE NEXT 6 SECS. CALCULATE THE DISTANCE

PLEASE PLEASE HELP

- PHYSICS -
**Elena**, Friday, June 8, 2012 at 3:07pm
v=a1•t => a=v/t =36/4 = 9 m/s²

s=at²/2=9•4²/2 =72 m.

v1=v-a1•t1

a1=(v-v1)/t1=(36-32)/6 =0.67.

s1=v•t1-a1•t1²/2 =36•6-(0.67•0.67)/2 =215.9 m.

S = s+s1 =72+215.9 = 287.9 m

- PHYSICS -
**Elena**, Friday, June 8, 2012 at 3:37pm
Mistake!!!

s1=v•t1-a1•t1²/2 =

=36•6-(0.67•6²)/2 =203.94 m.

S = s+s1 =72+203.94 = 275.94 m

- PHYSICS -
**MathMate**, Friday, June 8, 2012 at 4:47pm
It is more accurate to work with fractions. The above calculations are correct except for minor rounding errors.

We also have to assume that the car started *from rest*, which was unfortunately not mentioned in the question.

If (2/3) had been used in place of 0.67, we would have got total distance = 276 m

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