If acceleration is -9.8m/s^-2, what is the velocity? Do I intergrate or derive?

Answer 1

You integrate to get the velocity.

position function f(x)

f ' (x) : first derivative is velocity

f '' (x) : second derivative is acceleration

So if you have the acceleration, you have to integrate it to get the velocity

V(t) = -9.8t + c

Answer 2

There are three concepts in physics/calculus that are closely linked.

i) Position

ii) Velocity

iii) Acceleration

Position may be defined as distance from an original with respect to a coordinate system as a function of time. Say, position is represented by x(t), where t represents time.

Velocity is defined as the instantaneous change in position with respect to time. In other words, velocity is the limit of the change of position over the change in time as the time interval approaches 0. This is exactly the derivative of position with respect to time, or x'(t).

Acceleration is defined as the instantaneous change in velocity with respect to time. This is exactly the derivative of velocity with respect to time, or x''(t).

To answer your question, you integrate acceleration to obtain velocity and derive velocity to obtain acceleration. Remember the definition of acceleration as an instantaneous change in velocity to help :)

If acceleration is -9.8 m/s^2, then velocity is -9.8t + c m/s, where c is a constant that represents information about initial velocity.

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