Question Video: Finding the Velocity of a Particle at a Given Time and the Time Interval during Which the Velocity is Decreasing | Nagwa Question Video: Finding the Velocity of a Particle at a Given Time and the Time Interval during Which the Velocity is Decreasing | Nagwa

# Question Video: Finding the Velocity of a Particle at a Given Time and the Time Interval during Which the Velocity is Decreasing Mathematics • Third Year of Secondary School

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A particle is moving in a straight line such that its displacement π meters after π‘ seconds is given by π = 4π‘Β³ β 55π‘Β² + 208π‘. Find the velocity of the particle at π‘ = 8 seconds. Find the time interval during which the velocity of the particle is decreasing.

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### Video Transcript

A particle is moving in a straight line such that its displacement π meters after π‘ seconds is given by π is equal to four π‘ cubed minus 55π‘ squared plus 208π‘. Find the velocity of the particle at π‘ equals eight seconds. Find the time interval during which the velocity of the particle is decreasing.

The displacement of the particle can be expressed as a function of time π of π‘. And in this question, this is equal to four π‘ cubed minus 55π‘ squared plus 208π‘. We recall that the derivative of π of π‘ with respect to π‘ is π£ of π‘. And itβs an expression for the velocity in terms of time that we need to work out in the first part of this question.

We will do this by differentiating our function π of π‘ term by term. Using the power rule of differentiation, the derivative of four π‘ cubed is 12π‘ squared. Differentiating negative 55π‘ squared gives us negative 110π‘. And finally, the derivative of 208π‘ is the constant 208. π£ of π‘ is equal to 12π‘ squared minus 110π‘ plus 208.

We are asked to find the velocity of the particle at π‘ equals eight seconds. Substituting π‘ equals eight into our expression, we have π£ of eight is equal to 12 multiplied by eight squared minus 110 multiplied by eight plus 208. This simplifies to 768 minus 880 plus 208, which is equal to 96. Since the displacement was measured in meters and time in seconds, our units for velocity are meters per second. The velocity of the particle at π‘ equals eight seconds is 96 meters per second.

Letβs now consider the second part of the question. We are asked to find the time interval during which the velocity of the particle is decreasing. In order to answer this, we need to study the sign of the derivative of π£ of π‘, that is, the acceleration function π of π‘. Since acceleration is the derivative of the velocity with respect to time, we can find an expression for π of π‘ by differentiating π£ of π‘ term by term.

Differentiating 12π‘ squared using the power rule of differentiation gives us 24π‘. Differentiating negative 110π‘ gives us the constant negative 110. Since differentiating a constant gives us zero, π of π‘ is equal to 24π‘ minus 110. This is a linear function with a positive slope. And we can therefore conclude that the acceleration is increasing. This means that in order to find the time interval during which the velocity is decreasing, we need to solve the inequality π of π‘ is less than zero.

We have 24π‘ minus 110 is less than zero. Adding 110 to both sides, 24π‘ is less than 110. Dividing through by 24, π‘ is less than 110 over 24, which simplifies to 55 over 12. Since time cannot be negative, the acceleration is therefore negative when π‘ is greater than or equal to zero and less than 55 over 12 seconds, which in turn means that the velocity of the particle is decreasing for values of π‘ in this interval.

Our two answers are 96 meters per second and π‘ is greater than or equal to zero and less than 55 over 12.

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