# Video: Finding the Magnitude of the Acceleration If the Velocity Is Uniform

A particle is moving in a straight line such that its displacement 𝑠 in metres is given as a function of time 𝑡 in seconds by 𝑠 = 5𝑡³ − 84𝑡² + 33𝑡, 𝑡 ≥ 0. Find the magnitude of the acceleration of the particle when the velocity is zero.

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### Video Transcript

A particle is moving in a straight line such that its displacement 𝑠 in metres is given as a function of time 𝑡 in seconds by 𝑠 equals five 𝑡 cubed minus 84𝑡 squared plus 33𝑡, for 𝑡 is greater than or equal to zero. Find the magnitude of the acceleration of the particle when the velocity is zero.

Here, we have a function for the displacement of a particle at 𝑡 seconds. We recall that, given our function for 𝑠 in terms of 𝑡, 𝑣 can be obtained by differentiating this with respect to 𝑡. And then 𝑎, the acceleration, can be found by differentiating again. That’s d𝑣 by d𝑡 or, alternatively, the second derivative of 𝑠 with respect to 𝑡. So initially, we’re just going to find the second derivative of our function 𝑠. We can find the first derivative by differentiating each term in turn. That’s three times five 𝑡 squared minus two times 84𝑡 plus 33. That’s 15𝑡 squared minus 168𝑡 plus 33.

We then differentiate once again to find an expression for the acceleration. That’s two times 15𝑡 minus 168 which simplifies to 30𝑡 minus 168. And we now have expressions for the velocity and the acceleration at time 𝑡 seconds. So what now? We’re looking to find the magnitude of the acceleration when velocity is equal to zero. So let’s set our equation for the velocity equal to zero and solve for 𝑡. We can solve this by factoring. And when we do we see the five 𝑡 minus one times three 𝑡 minus 33 equals zero. For this statement to be true, either five 𝑡 minus one must be equal to zero or three 𝑡 minus 33 must be equal to zero. And if we solve each of these for 𝑡, we see that 𝑡 is equal to one-fifth or 𝑡 is equal to 11. And the velocity is equal to zero at these times.

Let’s see what happens when we substitute these values for 𝑡 into the equation for acceleration. When 𝑡 is equal to a fifth, acceleration is equal to 30 times one-fifth minus 168 which is negative 162. When 𝑡 is equal to 11, the acceleration is 30 times 11 minus 168 which, this time, is 162. Remember though we were asked to find the magnitude of the acceleration; that’s the size. So here the negative sign is irrelevant. We can say then that the magnitude of the acceleration when 𝑣 is equal to zero is 162 metres per square second.