Asked by fahd
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
PLEASE PLEASE HELP
Answers
Answered by
Elena
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
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
Answered by
Elena
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
s1=v•t1-a1•t1²/2 =
=36•6-(0.67•6²)/2 =203.94 m.
S = s+s1 =72+203.94 = 275.94 m
Answered by
MathMate
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 <i>from rest</i>, 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
We also have to assume that the car started <i>from rest</i>, 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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