Describe motion of an object from the following scenarios.

Question

1. d=5+2t
2. d=5t+2t^2
3. When t=0,d=0,v=15, and a constant acceleration of -5.

Answer 1

For problem 1, since the distance is a function of the time, using d(t) = 5 + 2t, we can say the distance starts at "5" and increases at a rate or 2.0, in other words, it starts 5 units away from the "zero" position and travels at a velocity of 2 (we don't know the units of measure). We can also say that the acceleration is zero since there is no change in velocity (it's always 2).

For problem 2, we are also told the distance is a function of time using d(t) = 5t + 2t² as our function. That means we start at the "zero" position, but this time the velocity changes as we go. The velocity is the derivative of the position function meaning v(t) = 5 + 4t. We can also say, since the acceleration is the derivative of the velocity function, that the acceleration is 4 (again, we don't know the units).

For problem 3, we are told that the acceleration is a constant -5. That means, since acceleration is the derivative of velocity, the velocity function that has a derivative would be v(t) = -5t + (some constant). Since we are told that at t = 0, the velocity is 15, the constant must be 15. so v(t) = -5t + 15.

Since velocity is the derivative of the position function, we need to find a position function whose derivative is -5t + 15. That means position (call it d(t)) would be d(t) = -2.5t² + 15t + (a constant). Since we are told that at time t = 0, the position is zero, then the constant must be zero, therefore d(t) = -2.5t² + 15t

These all mean that the object starts at the "zero" position having a velocity of 15 but it is slowing down having a constant acceleration of -5. You can map out the position and velocity as a function of time using the equations.