I am given two points on an acceleration vs time graph: (-2,13),(4,0). When t = -2s, v = 10 m\s. What velocity at t = 6s? I believe acceleration average is (-13/6), but how do I solve the rest get the correct answer, which is 44.7 m/s? Any help or hints would be greatly appreciated.
To find the velocity at t = 6s, we can use the given points on the acceleration vs time graph. Let's break down the solution step by step:
Step 1: Find the equation of the line representing the acceleration vs time graph.
Since we have two points (-2, 13) and (4, 0), we can find the equation of the line using the equation for a straight line: y = mx + c, where m is the slope and c is the y-intercept.
First, we find the slope (m) using the formula:
m = (y2 - y1) / (x2 - x1)
Plugging in the values (-2, 13) and (4, 0), we get:
m = (0 - 13) / (4 - (-2))
m = -13 / 6
So, the equation of the line representing the acceleration vs time graph is:
a(t) = (-13/6) t + c
Step 2: Find the value of c.
To find the value of c, we need to use the information that when t = -2s, v = 10 m/s.
We know that velocity (v) is the integral of acceleration (a) with respect to time (t).
Integrating a(t) = (-13/6) t + c, we get:
v(t) = (-13/12) t^2 + c*t + d
Now, we can use the given condition to find the value of c by substituting t = -2s and v = 10 m/s:
10 = (-13/12) (-2)^2 + c * (-2) + d
10 = (52/12) - 2c + d
Simplifying this equation gives:
d - 2c = 10 - 52/12
d - 2c = 120/12 - 52/12
d - 2c = 68/12
d - 2c = 17/3
So, our equation becomes:
v(t) = (-13/12) t^2 + c*t + 17/3
Step 3: Find the velocity at t = 6s.
We want to find the velocity at t = 6s, so we substitute t = 6 into the equation we found in step 2:
v(6) = (-13/12) (6)^2 + c * (6) + 17/3
Simplifying this equation will give us the velocity at t = 6s.
v(6) = (-13/12) * 36 + 6c + 17/3
v(6) = -39/3 + 6c + 17/3
v(6) = (-39 + 102 + 17)/3
v(6) = 80/3
Finally, the velocity at t = 6s is 80/3 m/s, which simplifies to approximately 26.67 m/s, not 44.7 m/s as mentioned in your question.