A particle moves in a straight line such that at time 𝑡 seconds its displacement from a fixed point on the line is given by 𝑠 equals 𝑡 cubed minus 𝑡 squared minus three metres, when 𝑡 is greater than or equal to zero. Determine whether the particle is accelerating or decelerating when 𝑡 equals two seconds.
Here, we have a function for the displacement of a particle at 𝑡 seconds. To find an equation for the acceleration, we’re going to need to find the second derivative of this function. Let’s begin by finding the first derivative d𝑠 by d𝑡 which represents the velocity at 𝑡 seconds. The derivative of 𝑠 with respect to 𝑡 is three 𝑡 squared minus two 𝑡. We differentiate once again to find the acceleration. And we see that the acceleration at 𝑡 seconds is six 𝑡 minus two. We want to find the nature of the acceleration at 𝑡 equals two seconds. In other words, is it accelerating or decelerating? We’ll substitute 𝑡 equals two into the equation for the acceleration. That’s six times two minus two which is 10.
By itself this isn’t quite enough to establish whether the particle is accelerating or decelerating since we don’t know what direction it’s moving in. So we’re going to evaluate the velocity at 𝑡 equals two seconds. That’s three times two squared minus two times two which is eight. Since both velocity and acceleration are positive at 𝑡 equals two seconds, we know they’re acting in the same direction. So the particle itself must be accelerating when 𝑡 equals two.