In one dimension, a particle's position as a function of time is given by x(t) = (1.12 m/s) t - (6.7 m/s2) t2. What is the magnitude of the acceleration of the particle in units of m/s2?

Answer 1

The fundamental question is this: “what is acceleration”?

Acceleration is defined to be the rate at which velocity changes over time. Velocity, in turn, is defined as being the rate at which position changes over time.

So, the units of velocity are distance / time. In your case, the units of velocity are specifically meters per second (distance in meters / time in seconds).

The units of acceleration are then defined as velocity / time, or equivalently, (distance / time) / time, which can also be written as distance / time^2. In your case, this is meters per second squared.

Now, you are given an “equation of motion”, which defines the position of the particle as a function of time:

x(t) = 1.12 t - 6.7 t^2 .

This has the form x(t) = v_0*t + 1/2 a*t^2, where v_0 is the initial velocity (the velocity at t=0, before any acceleration has changed the velocity of the particle), and “a” is the acceleration, which changes the velocity of the particle over time.

Therefore, the magnitude of the acceleration of the particle is found by equating the value in the given equation with the corresponding term in the kinematic equation: 1/2 |a| = 6.7. Thus the acceleration magnitude is |a|=13.4

Hope this helps,

Sam