can somebody shows me how to get to the answer key-- a)=2.51s, b)=19.9m, and c)=11.8i-16.1j A 1.57-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) = (7.00ti − 8.00j)N with t in seconds.
(a) At what time t will the particle's speed be 19.0 m/s?
s
(b) How far from the origin will the particle be when its velocity is 19.0 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 answers, we need to use the given information about the particle's force as a function of time, F(t) = (7.00ti − 8.00j) N, and the initial conditions of the particle being at rest at the origin.
(a) To find the time at which the particle's speed is 19.0 m/s, we first need to find the particle's acceleration. The acceleration is given by Newton's second law, F = ma, where F is the force and a is the acceleration.
Given F(t) = (7.00ti − 8.00j) N, we can equate it to ma:
ma = 7.00ti − 8.00j
Since the particle is initially at rest (v = 0), we can integrate the acceleration to find the velocity at time t:
∫(dv/dt) dt = ∫[(7.00ti − 8.00j)/m] dt
Integrating, we get:
v = (7.00/2)m*t^2 - 8.00jt + C
Since the particle is initially at rest (v = 0), the constant C is equal to 0. Therefore, we have:
v = (7.00/2)m*t^2 - 8.00jt
To find the time at which the particle's speed is 19.0 m/s, we set v = 19.0 m/s:
19.0 = (7.00/2)m*t^2 - 8.00jt
Now, we solve this equation for t. Rearranging the equation, we have:
(7.00/2)m*t^2 - 8.00jt - 19.0 = 0
This is a quadratic equation in t. By solving this equation, you can find the time t at which the particle's speed is 19.0 m/s.
(b) To find the distance from the origin when the velocity is 19.0 m/s, we can use the equation for displacement, s = ∫v(t) dt.
Since velocity is the derivative of displacement, integrating the velocity function will give us the displacement s.
Integrating v = (7.00/2)m*t^2 - 8.00jt:
s = (7.00/6)m*t^3 - 8.00jt^2 + C
Again, since the particle starts at the origin (s = 0), the constant C is 0. Therefore, the equation for displacement becomes:
s = (7.00/6)m*t^3 - 8.00jt^2
To find the distance from the origin, we need to find s when the velocity is 19.0 m/s. Substitute v = 19.0 m/s into the equation and solve for t to find the time at which the velocity is 19.0 m/s, then substitute this value of t into the equation to find the distance from the origin.
(c) The total displacement of the particle at this time can be calculated by setting up another integral of velocity with respect to time. The displacement is the integral of the velocity function over the specific time interval.
The displacement, r, is given by:
r = ∫v(t) dt
Integrating v = (7.00/2)m*t^2 - 8.00jt:
r = (7.00/6)m*t^3 - 8.00jt^2 + C
Again, since the particle starts at the origin (r = 0), the constant C is 0. Therefore, the equation for displacement becomes:
r = (7.00/6)m*t^3 - 8.00jt^2
To find the total displacement, substitute the value of t at which the velocity is 19.0 m/s and evaluate the expression for r. The result will give you the vector form of the total displacement.