An object moves with acceleration 350 ms2 and over a time in

An object moves with acceleration 3.50 m/s^2 and over a time interval reaches a final velocity of 12.8 m/s. If its initial velocity is 6.4 m/s, what is its displacement during the time interval? What is the distance it travels during this interval? If its initial velocity is -6.4 m/s, what is its displacement during the time interval? What it the total distance it travels during the interval in part (c)?

Solution

a)
v^2=u^2+2as
s=(v^2-u^2)/(2a)
=(12.8^2-6.4^2)/(2*3.5)
=17.55 m

b)
17.55m, as the acceleration is in the same direction with the velocity

c)
v^2=u^2+2as
s=(v^2-u^2)/(2a)
=(12.8^2-(-6.4)^2)/(2*3.5)
=17.55 m

d)
v^2=u^2+2as
s1=(v^2-u^2)/(2a)
=(0^2-(-6.4)^2)/(2*3.5)
=5.85 m

s2=(v^2-u^2)/(2a)
=(12.8^2-0^2)/(2*3.5)
=23.40 m
s= s1+s2
=5.85+23.40
=29.25 m