A paperweight is thrown straight up with an initial upward velocity of 60 meters per second and an initial height of 6 meters. The acceleration (due to gravity) of the object is given by a(t) = - 9.8 meters per second, where t is in seconds. The equation that describes the paperweight's velocity as a function of time t is:
I’m am sort of stuck on this problem, i am thinking of taking the antiderivative of the acceleration for velocity and then plus the 60 (initial upward velocity). Which turns out to be -9.8t+60+C
But I’m not feeling confident about this answer. The initial height wasn’t used as part of my solution and there should be a way to find C. I am not sure how.
Thank you so much ☺️
Its 189.673
Because yes, like oobleck said:
a(t) = -9.81
v(t) = -9.81t + C
since v(0) = 60,
v(t) = -9.81t + 60
h(t) = -4.9t^2 + 60t + C
since h(0) = 6,
h(t) = -4.9t^2 + 60t + 6
Velocity is 0 which means -9.8t+60=0 which means t=60/9.8
a(t) = -9.81
v(t) = -9.81t + C
since v(0) = 60,
v(t) = -9.81t + 60
h(t) = -4.9t^2 + 60t + C
since h(0) = 6,
h(t) = -4.9t^2 + 60t + 6
Stone
Well, don't get yourself in knots over this problem! Let me help you unravel the mystery of finding the velocity equation.
You're on the right track with taking the antiderivative of the acceleration to find the velocity equation. So, integrating the acceleration function "a(t) = -9.8" gives you "-9.8t".
Now, to find C, let's use the initial condition given in the problem: the paperweight has an initial upward velocity of 60 m/s. Since velocity is the derivative of position, we can set the derivative of the position function to the given initial velocity.
The position function, s(t), can be obtained by integrating the velocity function, which is "-9.8t + C". So, integrating "-9.8t + C" gives you "-4.9t^2 + Ct + D", where D is another constant of integration.
The initial height of 6 meters gives us s(0) = 6. Substituting t = 0 and s(0) = 6 into the position function, we can solve for D. That's one mystery solved!
Now, to find C, we can use the initial velocity condition, which gives us v(0) = 60. Differentiating the position function, we get the velocity function "v(t) = -9.8t + C". Substituting t = 0 and v(0) = 60 into the velocity function, we can solve for C.
Once you have solved for C, you can confidently write the equation that describes the paperweight's velocity as a function of time t, which is "-9.8t + C".
So, keep your chin up, solve for C, and laugh in the face of confusion! You've got this, my friend.
You're on the right track! To find the equation that describes the paperweight's velocity as a function of time, let's go step by step.
Step 1: Integrate the acceleration function, a(t), to find the velocity function, v(t).
Since a(t) is given as -9.8 m/s², in order to integrate it with respect to time, you can simply multiply it by the variable t:
∫(-9.8) dt = -9.8t + C₁
Here, C₁ is the constant of integration.
Step 2: Add the initial velocity, v₀, to the velocity function.
Given that the initial upward velocity is 60 m/s, you can add this initial velocity to the equation derived in step 1:
v(t) = -9.8t + C₁ + v₀
Plugging in the known values, you obtain:
v(t) = -9.8t + C₁ + 60
Step 3: Use the initial height to find the value of the constant of integration, C₁.
The initial height of the papwerweight is given as 6 meters. When the paperweight is at its highest point (at maximum height), its velocity is zero. Using this information, you can set v(t) = 0 and solve for t:
-9.8t + C₁ + 60 = 0
From this equation, you can solve for C₁:
C₁ = -60 + 9.8t
However, we still need to find the value of t when v(t) = 0. So, let's set:
-9.8t + C₁ + 60 = 0
Substituting the value of C₁, we have:
-9.8t - 60 + 9.8t = 0
Simplifying, we get:
-60 = 0
This is not a valid solution, indicating an error. Therefore, we need to revisit the problem setup or the equations provided.
It's possible that there is an error in the problem statement or that some additional information is required. Please double-check your equations and any given information to ensure accuracy.