A brick mason is on a steep roof repairing a chimney and drops a brick. If the brick slides down the roof and leaves the edge of the roof with the velocity
v = (1.00 m/s)î − [(6.00 m/s) + (9.80 m/s2)t]ĵ,
determine the displacement of the brick from the edge of the roof at a time
t = 0.330 s
after it leaves the edge of the roof. (Express your answer in vector form.)
Ah, the adventures of a wayward brick on a slippery roof! Let's see where it ends up.
To find the displacement, we need to integrate the velocity with respect to time. So let's grab our calculators and get to work!
Integrating the x-component, we get:
∫(1.00 m/s) dt = (1.00 m/s)(0.330 s) = 0.33 m
Integrating the y-component, we get:
∫[-(6.00 m/s) - (9.80 m/s^2)t] dt = [-6.00t - (9.80/2)t^2] evaluated from 0 to 0.330 s
Plugging in the values, we get:
[-6.00(0.330) - (9.80/2)(0.330)^2] - [-6.00(0) - (9.80/2)(0)^2] = -1.98 m
Therefore, the displacement of the brick from the edge of the roof at t = 0.330 s is:
Displacement = (0.33 m)î + (-1.98 m)ĵ
So, the wayward brick ended up about 0.33 meters horizontally and 1.98 meters vertically away from the edge of the roof. Ah, the perils of gravity and slippery slopes!
To determine the displacement of the brick from the edge of the roof at a time t = 0.330 s, we need to integrate the velocity vector with respect to time.
Given that the velocity vector is v = (1.00 m/s)î - [(6.00 m/s) + (9.80 m/s^2)t]ĵ, we can integrate each component separately.
Integrating the x-component:
∫(1.00 m/s) dt = (1.00 m/s) t + C₁
where C₁ is the constant of integration for the x-component.
Integrating the y-component:
∫[(6.00 m/s) + (9.80 m/s^2)t] dt = (6.00 m/s)t + (4.90 m/s^2)t^2 + C₂
where C₂ is the constant of integration for the y-component.
Now, plugging in t = 0.330 s into the expressions above, we can find the constants of integration:
For the x-component, we have:
C₁ = (1.00 m/s)(0.330 s) = 0.33 m
For the y-component, we have:
C₂ = (6.00 m/s)(0.330 s) + (4.90 m/s^2)(0.330 s)^2 = 1.98 m + 0.543 m = 2.52 m
So, the displacement vector at t = 0.330 s is given by:
d = (1.00 m/s)(0.330 s)î + (0.33 m - [(6.00 m/s)(0.330 s) + (4.90 m/s^2)(0.330 s)^2])ĵ
= (0.33 m)î + (0.33 m - 2.52 m)ĵ
= (0.33 m)î - (2.19 m)ĵ
Therefore, the displacement of the brick from the edge of the roof at t = 0.330 s is (0.33 m)î - (2.19 m)ĵ.
To determine the displacement of the brick from the edge of the roof at a specific time, we need to integrate the equation of its velocity with respect to time.
Given:
v = (1.00 m/s)i - [(6.00 m/s) + (9.80 m/s^2)t]j
t = 0.330 s
To find the displacement, we integrate the velocity equation with respect to time from the initial time (t = 0) to the given time (t = 0.330 s).
∫[(1.00 m/s)i - [(6.00 m/s) + (9.80 m/s^2)t]j] dt
Taking the integral of the i-component:
∫(1.00 m/s) dt = (1.00 m/s) t + Ci
Taking the integral of the j-component:
∫[(6.00 m/s) + (9.80 m/s^2)t] dt = (6.00 m/s)t + (4.90 m/s^2)t^2 + Cj
Substituting the given time t = 0.330 s into the equations:
Displacement in the x-direction (i-component):
(1.00 m/s) (0.330 s) = 0.330 m
Displacement in the y-direction (j-component):
(6.00 m/s) (0.330 s) + (4.90 m/s^2) (0.330 s)^2 = 0.8745 m + 0.544 m = 1.4185 m
Therefore, the displacement of the brick from the edge of the roof at t = 0.330 s is:
Displacement = (0.330 m)i + (1.4185 m)j