Question Video: Finding the Maximum Velocity of a Particle | Nagwa Question Video: Finding the Maximum Velocity of a Particle | Nagwa

# Question Video: Finding the Maximum Velocity of a Particle Mathematics • Third Year of Secondary School

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A particle is moving in a straight line along the 𝑥-axis such that its displacement 𝑠 meters after 𝑡 seconds is given by 𝑠(𝑡) = −𝑡³ + 6𝑡² + 2𝑡. Find the maximum velocity of the particle in the positive 𝑥-direction.

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### Video Transcript

A particle is moving in a straight line along the 𝑥-axis such that its displacement 𝑠 meters after 𝑡 seconds is given by 𝑠 of 𝑡 is equal to negative 𝑡 cubed plus six 𝑡 squared plus two 𝑡. Find the maximum velocity of the particle in the positive 𝑥-direction.

In this question, the displacement of the particle is expressed as a function of time. This is given by 𝑠 of 𝑡 is equal to negative 𝑡 cubed plus six 𝑡 squared plus two 𝑡, where 𝑡 is greater than or equal to zero. We know that the derivative of 𝑠 of 𝑡 with respect to 𝑡 is 𝑣 of 𝑡. This means that we can find an expression for the velocity in terms of time by differentiating term by term. Using the power rule of differentiation, 𝑣 of 𝑡 is equal to negative three 𝑡 squared plus 12𝑡 plus two. And since the displacement was measured in meters and time in seconds, the velocity has units of meters per second.

This velocity function is a quadratic function with a negative leading coefficient negative three. Its graph is therefore a parabola opening downwards. And the vertex of the parabola corresponds to a maximum value of the velocity as the slope of the graph at this point and, hence, its acceleration equals zero. We can find this value by firstly calculating the derivative of the velocity, that is, the acceleration, and finding at what time it is zero. Once again, we differentiate term by term, giving us 𝑎 of 𝑡 is equal to negative six 𝑡 plus 12, noting that differentiating a constant gives us zero. Setting this function equal to zero, we have negative six 𝑡 plus 12 equals zero. We can then add six 𝑡 to both sides. And dividing through by six, we have 𝑡 is equal to two. The acceleration of the particle is equal to zero when 𝑡 is equal to two seconds.

We can now substitute this value of 𝑡 back into our expression for the velocity. 𝑣 of two is equal to negative three multiplied by two squared plus 12 multiplied by two plus two. This simplifies to negative 12 plus 24 plus two, which is equal to 14. This tells us that an extrema of the velocity is 14 meters per second. And whilst it appears that this will be the maximum velocity, it is worth substituting in some values to check.

We will check values of the velocity before and after 𝑡 equals two seconds. Substituting 𝑡 equals one into our expression for 𝑣 of 𝑡, we have 𝑣 of one is equal to 11 meters per second. Likewise, when 𝑡 is equal to three seconds, the velocity is also equal to 11 meters per second. As these are less than 14, this confirms that we have a maximum velocity at 𝑡 equals two seconds, which is equal to 14 meters per second.

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