A particle moves according to the equation x(t) = (10 m/s2) t2 where x is in meters and t is in seconds. Find the average velocity for the time interval from t1 = 2.0 s to t2 = 3 s. Answer in units of m/s.
Please help! I have no idea how to set this up :(
x(3) = 90m
x(2) = 40m
total distance: 50m
total time: 1s
avg speed: 50 m/s
To find the average velocity for the given time interval, you first need to find the displacement of the particle during this interval. Then, you can divide the displacement by the time elapsed to find the average velocity.
Given that the equation for the position of the particle is x(t) = (10 m/s^2) t^2, we can find the displacement by subtracting the initial position from the final position. Let's calculate the displacement first.
At time t1 = 2.0 s, we can substitute this value into the equation to find the initial position:
x(t1) = (10 m/s^2)(2.0 s)^2 = 40 m
Similarly, at time t2 = 3 s, the final position can be found:
x(t2) = (10 m/s^2)(3 s)^2 = 90 m
The displacement during the given time interval is the difference between these two positions:
Δx = x(t2) - x(t1) = 90 m - 40 m = 50 m
Now, to find the average velocity, we divide the displacement by the time elapsed:
average velocity = Δx / Δt
Where Δt is the time interval, which is t2 - t1 = 3 s - 2.0 s = 1.0 s.
Let's substitute the values into the formula:
average velocity = 50 m / 1.0 s = 50 m/s
Therefore, the average velocity for the time interval from t1 = 2.0 s to t2 = 3 s is 50 m/s.