Answer:
a) The displacement is -4.5 m.
b) The traveled distance is 11.7 m.
Explanation:
Hi there!
a)The velocity of the particle is the derivative of the displacement function, x(t):
v(t) = dx/dt = 5t - 9
Separating varibles:
dx = (5t - 9)dt
Integrating both sides from x = x0 to x and from t = 0 to t.
x - x0 = 1/2 · 5t² - 9t
x = 1/2 · 5t² - 9t + x0
If we place the origin of the system of reference at x = x0, the displacement at t = 3 will be x(3):
x(3) = 1/2 · 5 · (3)² - 9(3) + 0
x(3) = -4.5
The displacement at t = 3 s is -4.5 m. It means that the particle is located 4.5 m to the left from the origin of the system of reference at t = 3 s.
b) When the velocity is negative, the particle moves to the left. Let´s find the time at which the velocity is less than zero:
v = 5t - 9
0 > 5t - 9
9/5 > t
1.8 s > t
Then until t = 1.8 s, the particle moves to the left from the origin of the reference system.
Let´s find the position of the particle at that time:
x = 1/2 · 5t² - 9t
x = 1/2 · 5(1.8 s)² - 9(1.8 s)
x = -8.1 m
From t = 0 to t = 1.8 s the traveled distance is 8.1 m. After 1.8 s, the particle has positive velocity. It means that the particle is moving to the right, towards the origin. If at t = 3 the position of the particle is -4.5 m, then the traveled distance from x = -8.1 m to x = -4.5 m is (8.1 m - 4.5 m) 3.6 m.
Then, the total traveled distance is (8.1 m + 3.6 m) 11.7 m.
