a uniformly accelerated car is found to possess the velocities 12.5 and 135 kmph at two different places in its path . if the car takes 10 minutes to travel between these places. find the distance between them.
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Given is
initial velocity = 12.5km/h
final velocity = 135km/h
time = 10 min = 10÷60 = 1/6hr
so according to the formula
v = u + at
135 = 12.5 + a*1/6
135 - 12.5 = a/6
122.5 * 6= a. (on subtracting 135 & 12.5)
a = 735km/h^2
now we have to find the distance between them
so accorrding to the formula
S = ut + 1/2 at^2
S = 135*1/6 + 735*(1/6)^2
S = 22.5 + 20.42
S = 42.92km
according to the given values this is your answer.
initial velocity = 12.5km/h
final velocity = 135km/h
time = 10 min = 10÷60 = 1/6hr
so according to the formula
v = u + at
135 = 12.5 + a*1/6
135 - 12.5 = a/6
122.5 * 6= a. (on subtracting 135 & 12.5)
a = 735km/h^2
now we have to find the distance between them
so accorrding to the formula
S = ut + 1/2 at^2
S = 135*1/6 + 735*(1/6)^2
S = 22.5 + 20.42
S = 42.92km
according to the given values this is your answer.
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