If an object with uniform acceleration (or deceleration) has a speed of #3 m/s# at #t=0# and moves a total of 4 m by #t=5#, what was the object's rate of acceleration?

1 Answer
A08
Jan 13, 2016

#4= 3 * 5 + (1)/(2)a * 5^2#
Solving for a
#a= -(22)/25 (m) / sec^2#
Deceleration

Explanation:

Use the equation
#s = ut +(1)/(2)a t^2#
where s is the distance traveled
u is initial velocity at t = 0
a is acceleration
and t is time of travel
#-#ve sign means deceleration

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