Question

4. The equations below give the x component of position for four objects as functions of time: Object 1: x(t) = a, where a 5m . Object 2: x()btc, where b 4and c-1m Object 3: x(t) = eẹ+ft,'where e= +5st and f--g2 Object 4: x(t)gt2 h, where g-3and h +12m . a. b. c. Which objects have a velocity that changes with time? Which object is at the origin at the earliest instant, and what is that instant? For the object found in part b, what is its velocity 1 s after it is at the origin?

Answer 1

(a) We know that the velocity is given by
V = dx/dt
Now for Object 1
$V_{1} = \frac{dx}{dt} = \frac{da}{dt} = 0$
hence the velocity is zero or do not depend upon the time
Now for Object 2
$V_{2} = \frac{dx}{dt} = \frac{d(bt+c)}{dt} = b$
V₂ = 4 m/s
Hence the velocity of the object 2 is constant or not depend upon the time.
Now for object 3
$V_{3} = \frac{dx}{dt} = \frac{d(et^{2}+ft)}{dt} = 2et +f$
hence it will depend upon time.
Now for object 4
$V_{4} = \frac{dx}{dt} = \frac{d(gt^{2}+h)}{dt} = 2gt$
Therefore it will also depend upon time.
Hence, object 3 and 4 velocity will depend upon time.
(b)
We know that origin X = 0 , therefore
For Object 1
X(t) = a = 5 m
Hence it is independent of time and it will remain at 5 m .
For Object 2
x(t) = bt+c
putting the values
0 = 4t -1
t = 1/4 = 0.25 s
Therefore object 2 will be at origin at t = 0.25 s
Now for object 3 ,
x(t) = et² +ft
0 = 5t² + (-9)t
5t = 9
t = 9/5 = 1.8 s
Hence the object 3 will be at origin at t = 1.8 s.
Now for object 4
x(t) = gt² + h
0 = (-3)t² + 12
t² = 4
t = 2 s
hence the object 4 will be at origin at t = 2 s
Therefore object 2 will be at earliest on the origin.
(c)
We have calculated the velocity of object 2 in part 1
V₂ = 4 m/s
Hence its velocity after 1 s will be 4 m/s.