Instantaneous Acceleration?

Question

An object moves along the x-axis according to the equation x(t)=(2.50t^2 - 2.00t + 3.00) m.

What is the instantaneous acceleration of the object at:

a. t=2.3

b. t=4

Answer 1

Velocity is the derivative of x so v=5t-2

Acceleration is the derivative of velocity so in this case a=5.

Because the equation a=5 doesn't have a t in it, the acceleration would be 5 at any time. (If you wanted the velocity you would have to insert the given time into t)

Answer 2

position=

x=2.5t^2-2*t+3

velocity=

x'=5t-2

acceleration=

x''=5 m/s^2  "a constant"

a.) 5 m/s^2

b.) 5 m/s^2

time is not a factor

Answer 3

v(t) = x'(t) and a(t) = x''(t), so take the second derivative of x(t):

x'(t) = (5m/s^2)t + (2m/s) = v(t), and x"(t) = 5m/s^2 = a(t)

The acceleration remains constant as the power of the change in position is 2 (if it was a power of 3, the acceleration would change with the time, and so you'd have to plug in the time into the acceleration of t equation).

Without using derivatives, you can use the equations x(t) = (1/2)at^2 + v(initial)t + x(initial), and from this equation realize that 2.5 is 1/2 of a.

Answer 4

acceleration a(t)=5

therefore,

a) a = 5 m/s^2

b) a = 5 m/s^2

Answer 5

Without doing all the work for you...

Take the derivative on x(t) twice. Then simply plug in t=2.23 and t=4 to get you answers.

(Hint, it's the same answer for both times!)