Question

An object moves with constant acceleration 3.65 m/s2 and over a time interval reaches a final velocity of 10.4 m/s.
(a) If its initial velocity is 5.2 m/s, what is its displacement during the time interval?
m

(b) What is the distance it travels during this interval?
m

(c) If its initial velocity is
−5.2
m/s, what is its displacement during the time interval?
m

(d) What is the total distance it travels during the interval in part (c)?
m

Answer 1

Assume that positive x-axis directed to the right
(a)(b) v=vₒ+at
t=( v-vₒ)/a=(10.4-5.2)/3.65=1.42 s
displacement = distance= vₒt+at²/2=5.2•1.42+3.65•1.42²/2=
7.384+3.68 =11.064 m
(from origin point - to theright)

(c) v=-vₒ+at
t1=(0+vₒ)/a=5.2/3.65=1.42 s.
s1 =vₒ²/2a=5.2²/2•3.65=3.7 m (to the left)
v=a•t2
t2=v/a=10.4/3.65=2.89 s.
s2= at²/2=3.65•2.89²/2=15.23 m (to the right)

Distance=3.7 +15.23=18.93 m

Displacement =15.23-3.7=11.53 m