a police car going 15.0 m/s east increases its velocity to 35.0 m/s east over a distance of 125 m. what is the magnitude of its acceleration during the 125 m
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You can use the third equation of motion: v^2 = u^2 + 2as
u: initial velocity
v: final velocity
a: acceleration
s: distance covered over the observed period
substituting values,
35^2 = 15^2 + 2*a*125
35^2 -15^2 = 250a
(35 + 15)(35 - 15) = 250a
50 * 20 = 250a
20 = 5 a
a = 4 m/s^2 East.
u: initial velocity
v: final velocity
a: acceleration
s: distance covered over the observed period
substituting values,
35^2 = 15^2 + 2*a*125
35^2 -15^2 = 250a
(35 + 15)(35 - 15) = 250a
50 * 20 = 250a
20 = 5 a
a = 4 m/s^2 East.