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
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
3 answers
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
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
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
3 answers