A car travels along a straight road for 30 seconds starting at time t = 0. Its acceleration in ft/sec2 is given by the linear graph below for the time interval [0, 30]. At t = 0, the velocity of the car is 0 and its position is 10.
What is the velocity of the car when t = 6?
I can't show the graph, but the y intercept is at (0,10) and the x intercept is (10,0).
So if the equation of the graph is 10-t, then the velocity equation would be 10t-t^2/2+C and I got C to be 0. I got 42 ft/sec to be my answer for the velocity at t=6 by plugging 6 in for t. Is this all there is to it or is there something I'm missing regarding the position formula?
To find the velocity of the car when t = 6, you have correctly determined the equation for the velocity, which is 10t - t^2/2. By plugging in t = 6 into this equation, you correctly found the velocity to be 42 ft/sec.
However, it seems like you might be confusing the acceleration graph with the velocity equation. The equation for the acceleration of the car, based on the given graph, is not 10 - t. Instead, the acceleration graph represents a linear decrease from 10 ft/sec^2 (at t = 0) to 0 ft/sec^2 (at t = 10). Since the acceleration is constant during this time interval, the equation for the acceleration is simply -10 ft/sec^2 (negative because it is a decrease).
To find the position equation, you would need to integrate the velocity equation with respect to time. Integrating 10t - t^2/2 with respect to t would give you the position equation, which is 10t^2/2 - (t^3)/6 + C. However, in this specific problem, you are only asked for the velocity at t = 6, so there is no need to find the position equation or worry about the constant C.
In summary, you correctly determined the velocity equation and correctly found the velocity at t = 6. Just remember to separate the concepts of acceleration, velocity, and position, and use the appropriate equations for each.