Question Video: Finding the Minimum Velocity of a Particle Mathematics • Higher Education

A particle moves along the π‘₯-axis. When its displacement from the origin is 𝑠 m, its velocity is given by 𝑣 = √(1/(βˆ’14𝑠² + 98)) m/s. Find the particle’s minimum velocity.

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Video Transcript

A particle moves along the π‘₯-axis. When its displacement from the origin is 𝑠 meters, its velocity is given by 𝑣 is equal to the square root of one over negative 14𝑠 squared plus 98 meters per second. Find the particle’s minimum velocity.

In this question, we are asked to find the minimum velocity of a particle. And we know that this occurs when the acceleration π‘Ž is equal to zero. We are given an expression for the velocity 𝑣 of the particle in terms of its displacement 𝑠. And we know that the acceleration π‘Ž is equal to 𝑣 multiplied by d𝑣 by d𝑠. This means that our first step will be to find an expression for d𝑣 by d𝑠. We will differentiate the expression we have been given with respect to 𝑠.

Using our knowledge of exponents or indices, we can rewrite the square root of one over negative 14𝑠 squared plus 98 as negative 14𝑠 squared plus 98 raised to the power of negative one-half. We can then use the chain rule to differentiate this expression with respect to 𝑠. d𝑣 by d𝑠 is equal to negative one-half multiplied by negative 14𝑠 squared plus 98 raised to the power of negative three over two multiplied by negative 28𝑠. We obtain this by multiplying the expression in the parentheses by the exponent. We reduce the exponent by one and then multiply by the expression in the parentheses differentiated. Differentiating negative 14𝑠 squared plus 98 with respect to 𝑠 gives us negative 28𝑠. Simplifying this, d𝑣 by d𝑠 is equal to 14𝑠 multiplied by negative 14𝑠 squared plus 98 raised to the power of negative three over two.

To find an expression for the acceleration π‘Ž, we multiply d𝑣 by d𝑠 by 𝑣. This gives us the expression shown. Since we are multiplying negative 14𝑠 squared plus 98 raised to the power of negative three over two by negative 14𝑠 squared plus 98 raised to the power of negative one-half, we can add the exponents. Negative three over two plus negative one-half is equal to negative four over two or negative two. π‘Ž is therefore equal to 14𝑠 multiplied by negative 14𝑠 squared plus 98 raised to the power of negative two.

Once again, we can rewrite this by recalling the laws of negative exponents. π‘Ž is equal to 14𝑠 divided by negative 14𝑠 squared plus 98 all squared. As already mentioned, the minimum velocity occurs when the acceleration is equal to zero. Setting our expression for π‘Ž equal to zero, and since the denominator cannot equal zero, we have 14𝑠 equals zero. Dividing through by 14, we have 𝑠 is equal to zero. The particle’s minimum velocity occurs when the displacement is equal to zero.

This means that we can substitute 𝑠 is equal to zero into our expression for 𝑣. 𝑣 is therefore equal to the square root of one over 98. This can be rewritten as the square root of one over the square root of 98. Using our laws of radicals or surds, the square root of 98 can be rewritten as the square root of 49 multiplied by the square root of two. And since the square root of 49 is seven, this is equal to seven root two. This means that 𝑣 is equal to one over seven root two.

Finally, we can rationalize the denominator by multiplying the numerator and denominator by root two, giving us root two over 14. The particle’s minimum velocity is equal to root two over 14 meters per second.

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