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.)
r =
m
To find the time at which the particle's speed is 14.0 m/s, we can use the equation:
Speed = |velocity|
We know that:
Speed = 14.0 m/s
And we can find the velocity by integrating the force with respect to time:
v = ∫ F(t) dt
Given the force as:
F(t) = (4.00ti - 8.00j) N
Integrating the x-component:
∫ (4.00ti) dt = 2.00t²i
Integrating the y-component:
∫ (-8.00j) dt = -8.00tj
Therefore, the velocity vector is:
v = (2.00t²i - 8.00tj)
Applying the equation:
Speed = |velocity|
14.0 m/s = sqrt((2.00t²)² + (-8.00t)²)
Simplifying the equation:
14.0² = 4.00t⁴ + 64.00t²
196.0 = 4.00t⁴ + 64.00t²
Rearranging the equation:
4.00t⁴ + 64.00t² - 196.0 = 0
This is a quadratic equation in terms of t². We can solve it by substituting:
u = t²
Now the equation becomes:
4.00u² + 64.00u - 196.0 = 0
We can solve this equation for u using the quadratic formula:
u = (-b ± sqrt(b² - 4ac)) / (2a)
In this case:
a = 4.00
b = 64.00
c = -196.0
Substituting the values:
u = (-64.00 ± sqrt(64.00² - 4 * 4.00 * -196.0)) / (2 * 4.00)
u = (-64.00 ± sqrt(4096.00 + 3136.00)) / 8.00
u = (-64.00 ± sqrt(7232.00)) / 8.00
u ≈ (-64.00 ± 85.11) / 8.00
We have two possible values for u:
u₁ ≈ (-64.00 + 85.11) / 8.00 ≈ 2.64
u₂ ≈ (-64.00 - 85.11) / 8.00 ≈ -18.64
Since time cannot be negative, we discard the solution u₂ ≈ -18.64.
Now, we can solve for t by taking the square root of u₁:
t = sqrt(u₁) = sqrt(2.64)
Therefore, the time at which the particle's speed is 14.0 m/s is approximately:
t ≈ sqrt(2.64) seconds.