Question Video: Evaluating the Expression of Displacement in Terms of Time and a Particle’s Uniform Velocity to Find an Unknown Expression | Nagwa Question Video: Evaluating the Expression of Displacement in Terms of Time and a Particle’s Uniform Velocity to Find an Unknown Expression | Nagwa

# Question Video: Evaluating the Expression of Displacement in Terms of Time and a Particleβs Uniform Velocity to Find an Unknown Expression Mathematics • Third Year of Secondary School

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A particle is moving in a straight line such that its displacement in meters, π , after π‘ seconds is given by π  = β12π‘Β³ + 12π‘Β² β 3π‘ m. When the particleβs velocity is zero, its acceleration is π m/sΒ². Find all the possible values of π.

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### Video Transcript

A particle is moving in a straight line such that its displacement in meters, π , after π‘ seconds is given by π  is equal to negative 12π‘ cubed plus 12π‘ squared minus three π‘ meters. When the particleβs velocity is zero, its acceleration is π meters per second squared. Find all the possible values of π.

The question tells us that a particle is moving in a straight line and that its displacement, π , after π‘ seconds is given by the equation π  is equal to negative 12π‘ cubed plus 12π‘ squared minus three π‘ meters. Weβre also told when the particleβs velocity is zero, its acceleration is π meters per second squared. We need to find all the possible values of π. Since weβre given the displacement of our particle which is traveling in a straight line, we can use the fact that velocity is the rate of change with respect to time of displacement.

We can use this to find the velocity of our particle at the time π‘. Itβs the derivative of the displacement function with respect to π‘. And we can differentiate this by using the power rule for differentiation, which tells us the constants π and π. The derivative of π times π‘ to the πth power with respect to π‘ is equal to πππ‘ to the power of π minus one. We multiply by the exponent and then reduce the exponent by one. Applying this gives us negative 36π‘ squared plus 24π‘ minus three. And it might help us to think of negative three π‘ as negative three π‘ to the first power.

Since the question wants us to find possible values for the acceleration, we need to find the acceleration of our particle. And we can do this by using the fact that the acceleration is the rate of change of the velocity. So to find the acceleration of our particle at the time π‘, we need to differentiate the velocity of our particle with respect to π‘. This gives the derivative of negative 36π‘ squared plus 24π‘ minus three with respect to π‘. We can do this by using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. This gives us negative 72π‘ plus 24.

The question is asking us to find the possible values of the acceleration when the velocity is equal to zero. So thatβs when our velocity function π£ of π‘ is equal to zero. Since our velocity function is a quadratic, we can find all of the values of π‘ which give a velocity of zero by solving a quadratic equation. So we set our velocity equal to zero. This means that any solutions to the quadratic equation, negative 36π‘ squared plus 24π‘ minus three, will give us a velocity of zero. We can simplify our quadratic by taking out a factor of negative three. This gives us negative three times 12π‘ squared minus eight π‘ plus one. There are a few different ways of solving this quadratic equation. For example, we could use the quadratic formula. However, in this case, we can actually factor our quadratic.

We want to choose our coefficients of π‘ to multiply to give us 12, for example, six and two. We then see that 16 minus one multiplied by two π‘ minus one gives us 12π‘ squared minus eight π‘ plus one. And since weβre solving this equal to zero, one of our factors must be equal to zero. Solving each of our factors equal to zero gives us π‘ is equal to one-sixth or π‘ is equal to one-half. Itβs worth noting at this point that we should check both of our solutions are valid. In this case, since π‘ represents the time, we need to make sure that π‘ is greater than or equal to zero, which is true in both of these cases.

Now, the question wants us to find the acceleration of our particle at these two values of π‘. We can do this by substituting them into our formula for the acceleration. So we substitute π‘ is equal to one-sixth and π‘ is equal to one-half into our formula for the acceleration. Substituting π‘ is equal to one-sixth gives us negative 72 times one-sixth plus 24, which is just equal to 12. And substituting π‘ is equal to one-half gives us negative 72 times one-half plus 24, which is equal to negative 12. And since we showed that the only times where our particle has zero velocity was when π‘ is equal to one-sixth or π‘ is equal to one-half. These are the only possible values for the acceleration when the velocity is zero.

Therefore, weβve shown that the only possible values for the acceleration π are negative 12 and 12.

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