Answer to Question #4303 in Mechanics | Relativity for Mike
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?
(b) What is the distance it travels during this interval?
(c) If its initial velocity is
m/s, what is its displacement during the time interval?
(d) What is the total distance it travels during the interval in part (c)?
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