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Calculus 1 : Acceleration

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Example Questions

Example Question #492 : Calculus

The position of a given kite is defined by the equation $p(t)=t^{3}+4t^{2}-6t+1$ . What is the acceleration of the kite at t=1 ?

Possible Answers:

Correct answer:

Explanation:

Acceleration a(t) is defined as the second derivative of position p(t) , or a(t)=p''(t) .

Therefore, given

$p(t)=t^{3}+4t^{2}-6t+1$ and using the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$

where $n\neq0$ ,

$p'(t)=3t^{2}+8t-6$ and

a(t)=p''(t)=6t+8 .

Consequently, at t=1 ,

a(1)=p''(1)=6(1)+8=6+8=14 .

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Example Question #493 : Calculus

The position of a given car is defined by the equation $p(t)=2t^{3}+3t^{2}+5t+12$ . What is the acceleration of the car at t=3 ?

Possible Answers:

Correct answer:

Explanation:

Acceleration a(t) is defined as the second derivative of position p(t) , or a(t)=p''(t) .

Therefore, given

$p(t)=2t^{3}+3t^{2}+5t+12$ and using the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$

where $n\neq0$ ,

$p'(t)=6t^{2}+6t+5$ and

a(t)=p''(t)=12t+6 .

Consequently, at t=3 ,

a(t)=p''(t)=12(3)+6=36+6=42 .

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Example Question #493 : Spatial Calculus

A tube is drilled through the center of Earth and all the magma (and air) is sucked out. A rock dropped in this tube has a position given by the equation

$p(t)=3959\cos{\frac{t}{807}}$

where is the number of seconds since it was dropped and p(t) is its displacement from the center of Earth in miles.

What is this rock's acceleration with respect to position?

Possible Answers:

$a(t)=\frac{-3959}{807^2} \cdot p(t)$

$a(t)=\frac{1}{807^2} \cdot p(t)$

$a(t)=\frac{-3959}{807} \cdot p(t)$

$a(t)=\frac{-3959^2}{807^2} \cdot p(t)$

$a(t)=\frac{-1}{807^2} \cdot p(t)$

Correct answer:

$a(t)=\frac{-1}{807^2} \cdot p(t)$

Explanation:

If we take the derivative of the rock's position, we get its velocity with respect to time:

$v(t)=-\frac{3959}{807}\sin{\frac{t}{807}}$

Taking the derivative again gives us acceleration:

$a(t)=-\frac{3959}{807^2}\cos{\frac{t}{807}}$

Note, that a(t) is proportional to p(t) , we know that, for some constant ,

$a(t)=k\cdot p(t)$

and

$k=\frac{a(t)}{p(t)}=\frac{-\frac{3959}{807^2}\cos{\frac{t}{807}}}{3959\cos{\frac{t}{807}}} = \frac{-\frac{3959}{807^2}}{3959}\frac{\cos{\frac{t}{807}}}{\cos{\frac{t}{807}}}$

The cosines cancel:

$k = \frac{-\frac{3959}{807^2}}{3959}=\frac{-1}{807^2}$

Therefore,

$a(t)=\frac{-1}{807^2} \cdot p(t)$

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Example Question #494 : Calculus

A ball is thrown off of a roof. Find the rate at which it is accelerating given the position function.

p(t)=-4t^2+5t+80

Possible Answers:

a(t)=8

a(t)=-8

a(t)=5

a(t)=-4t^2

Correct answer:

a(t)=-8

Explanation:

Given the position of an object, we can find its acceleration by taking the derivative twice.

We will use the power rule to differentiate which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Therefore, applying the power rule to each term in the position function we find our velocity. Applying the power rule a second time will result in the acceleration function.

$\\p(t)=-4t^2+5t+80\\ \Rightarrow p'(t)=v(t)=-8t+5\\ \Rightarrow p''(t)=v'(t)=a(t)=-8$

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Example Question #495 : Calculus

Find the accelaration given the position function.

p(t)=432t+47621

Possible Answers:

a(t)=432

a(t)=47621

a(t)=0

a(t)=864

Correct answer:

a(t)=0

Explanation:

Given the position of an object, we can find its acceleration by taking the derivative twice.

We will use the power rule to differentiate which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Therefore, applying the power rule to each term in the position function we find our velocity. Applying the power rule a second time will result in the acceleration function. Remember the derivative of a constant is always zero.

$\\ p(t)=432t+47621\\ p'(t)=v(t)=432\\ p''(t)=a(t)=0$

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Example Question #496 : Spatial Calculus

$\noindent f(x)=\sin(x^2+4).$

Find f''(x) .

Possible Answers:

${}f''(x)=2\cos(x^2+4)-4x^2\sin(x^2+4)$

${}f''(x)=2\cos(x^2+4)+4x^2\sin(x^2+4)$

${}f''(x)=2\cos(x^2+4)-4x\sin(x^2+4)$

${}f''(x)=2\sin(x^2+4)-4x^2\cos(x^2+4)$

${}f''(x)=2\cos(x^2+4)+4x\sin(x^2+4)$

Correct answer:

${}f''(x)=2\cos(x^2+4)-4x^2\sin(x^2+4)$

Explanation:

In order to find f''(x) , we first we need to find f'(x) .

We find this by applying the chain rule and power rule.

Remember that the chain rule is $[f(g(x))]'=f'(g(x))\cdot g'(x)$ .

The power rule is $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Also recall that trigonometric functions have special derivatives.

$\\f(x)=sin(x)\rightarrow f'(x)=cos(x)\\ \\f(x)=cos(x)\rightarrow f'(x)=-sin(x)$

Applying these rules we get the following first derivative of the function.

$\noindent f(x)=\sin(x^2+4).$

$\noindent f'(x)=\cos(x^2+4)(2x)=2x\cos(x^2+4)$

We get f''(x) by applying the chain rule, product rule and power rule.

Remember that the product rule is

$[f(x)\cdot g(x)]'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$ .

Applying these rules to the first derivative of the function we are able to find the second derivative.

$\\f'(x)=2x\cos(x^2+4)\\f''(x)=2\cos(x^2+4)+(2x)\cdot (-\sin(x^2+4))2x \\f''(x)=2\cos(x^2+4)-4x^2\sin(x^2+4)$

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Example Question #496 : Calculus

A particle is constrained to move only on the -axis. Its position $\small x(t)$ can be defined as $\small kt^{2}-bt+w$ , where , , and $\small w$ are constants. Find , , and $\small w$ if , and . represents the velocity at and represents the acceleration at .

Possible Answers:

$\small k=0, b=10, w=0$

Not enough information

$\small k=0, b=-10, w=0$

$\small k=5/2,b=-10,w=0$

$\small k=5/2, b=10, w=0$

Correct answer:

$\small k=5/2,b=-10,w=0$

Explanation:

We know , so

$\small \small 0=k(0)^2-b(0)+w\therefore w=0$ .

We also know , but we need to take the derivative of before we can use this knowledge. To find the derivative we will use the power rule which states,

$y=x^n \rightarrow \frac{dy}{dx}=nx^{n-1}$ .

Thus our derivative becomes,

$\small {\frac{\mathrm{d} }{\mathrm{d} t}}{}(kt^2-bt+w)=2kt-b$ .

Now we can apply to get:

$2k(0)-b=10=-b \therefore b=-10$

Now we need to take the derivative of to get

: $\frac{\mathrm{d} }{\mathrm{d} t}(2kt-b)=2k$

Using we get

$2k=5 \therefore k=5/2$

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Example Question #497 : Calculus

A tennis player serves a ball with an inital velocity of $45\frac{m}{s}$ . The position equation is given as S(t)=-t^3+7t^2+45t+10 . Find the acceleration of the ball t=5 .

Possible Answers:

The ball is not accelerating.

$-1.6\frac{m}{s^2}$

$-16\frac{m}{s^2}$

$1.6\frac{m}{s^2}$

$16\frac{m}{s^2}$

Correct answer:

$-16\frac{m}{s^2}$

Explanation:

For this problem a position equation is given to us and we are asked to find the acceleration fo the ball at t=5 . Remember that velocity is the derivative of position with respect to time and acceleration is the derivative of velocity with respect to time. Therefore, by taking the derivative twice on the original position equation, we will be able to obtain the acceleration equation.

To take the derivative of this equation, we must use the power rule,

$\frac{d}{dx}x^n=nx^{n-1}$ .

We also must remember that the derivative of an constant is 0. The first derivative of the position equation yields

v(t)=-3t^2+14t+45 .

Taking the derivative of the velocity equation through the power rule yields

a(t)=-6t+14 .

This acceleration equation gives us the acceleartion of the ball at any given time t.

Therefore substituting in t=5 we find that the acceleration for the ball at t=5 is $-16\frac{m}{s^2}$ .

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Example Question #498 : Calculus

A boat's position in meters is given by the function $f(t)=\frac{1}{10}t^5 + 2t^3+t$ .

Find the boat's acceleration at t=3 .

Possible Answers:

63m/s^2

90m/s^2

60m/s^2

300m/s^2

120m/s^2

Correct answer:

90m/s^2

Explanation:

To find the function for acceleration, take the second derivative of the position function.

To differentiate use the power rule which states,

$f(x)=x^n\rightarrow f'(x)=nx^{n-1}$ .

$f(t)=\frac{1}{10}t^5 + 2t^3+t$

$f'(t)=\frac{1}{2}t^4 + 6t^2+1$

f''(t)=2t^3 + 12t

Then, evaluate the acceleration function at t=3s to get the answer.

f''(3)=2(3)^3 + 12(3)

f''(3)=90m/s^2

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Example Question #101 : Acceleration

Find f''(x) of

$f(x)=ln(\sin(x))$ .

Possible Answers:

$f''(x)=\tan^2(x)$

$f''(x)=\csc(x)$

$f''(x)=-\csc^2(x)$

$f''(x)=\sec^2(x)$

Correct answer:

$f''(x)=-\csc^2(x)$

Explanation:

In order to find f''(x) , we need to find f'(x) first using the definition of derivative for the natural log and sine which are,

$ln(u)\rightarrow \frac{1}{u}\cdot u'$ and $sin(x)\rightarrow cos(x)$ .

Therefore we get,

$f(x)=ln(\sin(x))$ .

$f'(x)=\frac{1}{\sin(x)}\cdot \cos(x)=\cot(x)$

This was found by using the chain rule.

Remember that the chain rule is

$[f(g(x))]'=f'(g(x))\cdot g'(x)$

Now we can find f''(x) .

$f''(x)=-\csc^2(x)$

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #492 : Calculus

Example Question #493 : Calculus

Example Question #493 : Spatial Calculus

Example Question #494 : Calculus

Example Question #495 : Calculus

Example Question #496 : Spatial Calculus

Example Question #496 : Calculus

Example Question #497 : Calculus

Example Question #498 : Calculus

Example Question #101 : Acceleration

All Calculus 1 Resources

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