Video: MATH-DYNA-2018-S1-Q01

MATH-DYNA-2018-S1-Q01

03:26

Video Transcript

The velocity of a body is given by 𝑣 is equal to three π‘₯ squared minus four π‘₯.Find the acceleration π‘Ž in metres per second squared at π‘₯ is equal to two metres.

We know that the displacement is given by π‘₯.The velocity is given by 𝑣.And the acceleration is given by π‘Ž.Now, we know that velocity, 𝑣, is change in displacement π‘₯ over time, which is 𝑑.And so we can write this as 𝑣 is equal to dπ‘₯ by d𝑑.Also, we know that acceleration, π‘Ž, is change in velocity, 𝑣, over time, 𝑑.And so we can write that π‘Ž is equal to d𝑣 by d𝑑.Now, this question is asking us to find the acceleration at π‘₯ is equal to two metres.So we need to find an equation for π‘Ž in terms of π‘₯.Now, we do have an equation for 𝑣 in terms of π‘₯.And we know that π‘Ž is d𝑣 by d𝑑.However, since our equation for 𝑣 is in terms of π‘₯, we cannot differentiate this directly with respect to 𝑑.Instead, we need to apply the chain rule to d𝑣 by d𝑑.

Now, the chain rule tells us that if we are differentiating 𝑦 with respect to π‘₯, where π‘₯ and 𝑦 are two variables, then we can also write this as d𝑦 by d𝑒 times d𝑒 by dπ‘₯, where 𝑒 is a variable which we can choose.And so if we apply the chain rule to d𝑣 by d𝑑, we can write d𝑣 by d𝑑 as d𝑣 by dπ‘₯ multiplied by dπ‘₯ by d𝑑.And since we know 𝑣 in terms of π‘₯, we’re able to find d𝑣 by dπ‘₯.And we know that 𝑣 is equal to dπ‘₯ by d𝑑.So we can substitute 𝑣 in for dπ‘₯ by d𝑑.This gives us that π‘Ž is equal to 𝑣 times by d𝑣 by dπ‘₯.So now, we simply substitute in the equation for 𝑣 given in the question.We obtain three π‘₯ squared minus four π‘₯ times d by dπ‘₯ of three π‘₯ squared minus four π‘₯.

Now, the first three π‘₯ squared minus four π‘₯ will remain the same.And we just need to differentiate the second three π‘₯ squared minus four π‘₯.Differentiating the three π‘₯ squared term, we multiply by the power, which is two, giving us six, and reduce the power by one to one to give us π‘₯ to the power of one or just π‘₯.So three π‘₯ squared differentiates to six π‘₯.Then, to differentiate the negative four π‘₯ term, we multiply by the power of π‘₯, which is just one since π‘₯ is equal to π‘₯ to the power of one.Then, negative four times one gives us just negative four.Then, we simply reduce the power of π‘₯ by one to give us π‘₯ to the power of zero.And π‘₯ to the power of zero is simply one.So we’re left with negative four.

Now, we have found an equation for π‘Ž in terms of π‘₯.We have that π‘Ž is equal to three π‘₯ squared minus four π‘₯ times six π‘₯ minus four.All that remains to do is to substitute in π‘₯ is equal to two metres.And we obtain that π‘Ž is equal to three times by two squared minus four times two times by six times two minus four, where it simplifies down to 12 minus eight times 12 minus four or four times eight.

This gives us the solution that π‘Ž is equal to 32 meters per second squared.And it’s important that we don’t forget the units here, especially since the question reminds us.

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