A 1.49-kg particle initially at rest and at the origin of an x-y coordinate system is subjected to a time-dependent force of
F(t) = (4.00ti − 8.00j) N
with t in seconds.
(a) At what time t will the particle's speed be 14.0 m/s?
s
(b) How far from the origin will the particle be when its velocity is 14.0 m/s?
m
(c) What is the particle's total displacement at this time? (Express your answer in vector form. Do not include units in your answer.)
doesn't this one help any?
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To find the time, at which the particle's speed is 14.0 m/s, we need to solve the following equation:
|velocity| = speed
The velocity of a particle is given by the derivative of its position with respect to time:
velocity = d(position)/dt
In this case, the time-dependent force F(t) is given, so we can use Newton's second law to relate the force and the acceleration of the particle:
F = mass * acceleration
Since the mass of the particle is given as 1.49 kg, we can equate the force given by F(t) to the mass times the acceleration:
(4.00ti - 8.00j) N = 1.49 kg * acceleration
We can equate the x and y components separately:
4.00ti = 1.49 kg * ax ... (1)
-8.00j = 1.49 kg * ay ... (2)
Let's solve equation (1) for ax:
ax = (4.00t) / 1.49 ... (3)
We can differentiate this equation with respect to time to get the x-component of the acceleration:
dvx/dt = 4.00 / 1.49 ... (4)
Now, integrate equation (4) to find the x-component of the velocity:
vx = (4.00 / 1.49) * t + Cx ... (5)
Here, Cx is the constant of integration, which represents the initial velocity in the x-direction. Since the particle is initially at rest, Cx = 0.
Taking the magnitude of the velocity, we have:
|vx| = (4.00 / 1.49) * t
At the time t when the particle's speed is 14.0 m/s, we can set |vx| = 14.0 m/s:
(4.00 / 1.49) * t = 14.0
Solving for t, we get:
t = (14.0 * 1.49) / 4.00
Therefore, the time at which the particle's speed is 14.0 m/s is t = 5.22 seconds.
To find how far from the origin the particle will be when its velocity is 14.0 m/s, we can integrate the x-component of the velocity with respect to time:
x = ∫(4.00 / 1.49) * t dt
x = (2.68 / 1.49) * t^2 + C
Since the particle is initially at the origin (x = 0 at t = 0), C = 0. Plugging in the value of t = 5.22 seconds, we get:
x = (2.68 / 1.49) * (5.22)^2
Therefore, the particle will be x = 26.8 m from the origin when its velocity is 14.0 m/s.
To find the particle's total displacement at this time, we can integrate its velocity vector with respect to time:
displacement = ∫velocity dt
Integrating equation (5), we have:
displacement = ∫[(4.00 / 1.49) * t)i - 8.00j] dt
displacement = [(2.68 / 1.49) * t^2)i - 8.00tj] + C
Again, since the particle is initially at rest (displacement = 0 at t = 0), the constant of integration C = 0. Plugging in the value of t = 5.22 seconds, we get:
displacement = [(2.68 / 1.49) * (5.22)^2)i - 8.00(5.22)j]
Therefore, the particle's total displacement at this time is [(2.68 / 1.49) * (5.22)^2)i - 8.00(5.22)j].