Answer to Question #4303 in Mechanics | Relativity for Mike

Question

An object moves with constant acceleration 3.40 m/s2 and over a time interval reaches a final velocity of 11.8 m/s.
(a) If its initial velocity is 5.9 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
−5.9
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)?

Answer 1

a=3.4 acceleration of an object
Vf=11.8 final velosity of an object
V0=5.9 initial velosity
a) if an object moves with constant acceleration we can find its velosity at any time moment using formula
V=V0+at
from that formula we can find t:
t=(Vf-V0)/a=(11.8-5.9)/3.4 =1.74 s - total time

Displasement S=S0+V0t+at^2/2
where S0- initial position, S0=0
So displacement S=5.9*1.74+3.4*1.74^2/2=10.2+5.15=15.25 m
distance is same thing as displacement here, because S>0 . In general distance is scalar quantity and displacement has direction(it's vector quantity)
b) distance& 15.25 m as explained above
c ) Now V0=-5.9
t=(Vf-V0)/a=(11.8+5.9)/2=5.2
Displacement
S=V0t+at^2/2=-5.9*5.2+3.4*5.2^2/2=-30.9+46=15.1 m
distance is same thing as displacement here, because S>0 . So distance D=15.1 m
d)distance D=15.1 m