# Video: Finding the Velocity When the Acceleration Is Uniform and given the Displacement Expression

A body moves along the 𝑥-axis such that at time 𝑡 seconds, its displacement from the origin is given by 𝑠 = (6𝑡⁴ − 𝑡³ − 3𝑡² − 4𝑡 + 3) m, 𝑡 ≥ 0 . What is its velocity when its acceleration is equal to 0?

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### Video Transcript

A body moves along the 𝑥-axis. Such that, at time 𝑡 seconds, its displacement from the origin is given by 𝑠 is equal to six 𝑡 to the fourth power. Minus 𝑡 cubed minus three 𝑡 squared minus four 𝑡 plus three metres. For values of 𝑡 where 𝑡 is greater than or equal to zero. What is its velocity when its acceleration is equal to zero?

The question tells us that the body is moving along the 𝑥-axis. And it gives us a function for its displacement from the origin at a time of 𝑡 seconds, which is a polynomial, in terms of metres. The question wants us to use this information to find the velocity of the particle when its acceleration is equal to zero.

We can start by asking the question, when will the body have an acceleration which is equal to zero? We’re given an expression to calculate the displacement from the origin of the body after 𝑡 seconds. And since the body is moving in a straight line, we know that, for a displacement 𝑠, the rate of change of displacement gives us the velocity. And the rate of change in velocity gives us the acceleration.

So to find the times when the body has an acceleration of zero, we need to find the second derivative of our displacement function with respect to 𝑡. We’ll start by finding the first derivative of 𝑠 with respect to 𝑡, which is of course also equal to the velocity of the body. This gives us the derivative of six 𝑡 to the fourth power minus 𝑡 cubed minus three 𝑡 squared minus four 𝑡 plus three with respect to 𝑡.

Since we’re differentiating a polynomial, we can differentiate this using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. This gives us the velocity of the body at the time 𝑡 is 24𝑡 cubed minus three 𝑡 squared minus six 𝑡 minus four.

We can now use this to find the acceleration of the body at the time 𝑡. It’s the second derivative of the displacement with respect to 𝑡, which is of course just equal to the rate of change of the velocity with respect to time. So the acceleration is equal to the derivative of 24𝑡 cubed minus three 𝑡 squared minus six 𝑡 minus four with respect to 𝑡.

We can differentiate this using the power rule for differentiation. And this gives us 72𝑡 squared minus six 𝑡 minus six. And we wanted to find when the acceleration was equal to zero. So we set our acceleration equal to zero. This means that zero is equal to 72𝑡 squared minus six 𝑡 minus six. And this is a quadratic equation. So we can solve for the values of 𝑡.

We can simplify the equation by dividing both sides of the equation by six. This gives us that zero is equal to 12𝑡 squared minus 𝑡 minus one. There’s a few different ways we could solve this quadratic equation. For example, we could use the quadratic formula. However, in this case, we can factor it fully.

We want our coefficients to 𝑡 to multiply to give us 12. For example, we could try four and three. Then if we want our constants to multiply to give us negative one. We can see that four 𝑡 plus one multiplied by three 𝑡 minus one gives us 12𝑡 squared minus 𝑡 minus one. Then this will be equal to zero when one of the factors is equal to zero. So we either have that 𝑡 is equal to negative a quarter or 𝑡 is equal to one-third.

However, in this case, 𝑡 represents the time. In fact, we’re told that 𝑡 must be greater than or equal to zero. So in fact, we only have one solution when 𝑡 is equal to one-third. So since the question wants us to find the velocity of the body when the acceleration is equal to zero. And we just showed the only time the acceleration is equal to zero is when 𝑡 is equal to one-third. We need to find 𝑣 evaluated at one-third.

Substituting 𝑡 is equal to one-third into the velocity gives us 24 times one-third cubed minus three times one-third squared minus six times one-third minus four. Which we can evaluate to give us negative 49 divided by nine. In fact, since we’re measuring the time in seconds and the displacement in metres, our velocity, which is the rate of change of displacement, must have the units metres per second.

Therefore, we’ve shown if a body moves along the 𝑥-axis such that, at a time of 𝑡 seconds, its displacement from the origin is given by 𝑠 is equal to six 𝑡 to the fourth power. Minus 𝑡 cubed minus three 𝑡 squared minus four 𝑡 plus three metres, where 𝑡 is greater than or equal to zero. Then its velocity when its acceleration is equal to zero is negative 49 divided by nine metres per second.