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Calculus 1 : Spatial Calculus

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Example Questions

Example Question #280 : How To Find Acceleration

The velocity of a particle is given by the function $\displaystyle v(t)=4sin(t^4)$. What is the acceleration of the particle at time t=3 $\displaystyle t=3$ ?

Possible Answers:

83.9 $\displaystyle 83.9$

-272.1 $\displaystyle -272.1$

38.9 $\displaystyle 38.9$

335.5 $\displaystyle 335.5$

-25.7 $\displaystyle -25.7$

Correct answer:

335.5 $\displaystyle 335.5$

Explanation:

Acceleration of a particle can be found by taking the derivative of the velocity function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of velocity with respect to time, we are evaluating how velocity changes over time; i.e acceleration! This is just like finding velocity by taking the derivative of the position function.

$a(t)=\frac{d}{dt}v(t)$ $\displaystyle a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

Taking the derivative of the function

$\displaystyle v(t)=4sin(t^4)$

We'll need to make use of the following derivative rule(s):

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$ $\displaystyle d[ln(u)]=\frac{du}{u}$

Trigonometric derivative:

$\displaystyle d[sin(u)]=cos(u)du$

Note that u may represent large functions, and not just individual variables!

The acceleration function is

$\displaystyle a(t)=16t^3cos(t^4)$

At time t=3 $\displaystyle t=3$

$\displaystyle a(3)=16(3)^3cos(3^4)=335.5$

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Example Question #672 : Spatial Calculus

The velocity of a particle is given by the function $\displaystyle v(t)=9ln(sin(t))$. What is the acceleration of the particle at time $t=\frac{\pi}{3}$ $\displaystyle t=\frac{\pi}{3}$ ?

Possible Answers:

$9\sqrt{3}$ $\displaystyle 9\sqrt{3}$

$\frac{\sqrt{3}}{3}$ $\displaystyle \frac{\sqrt{3}}{3}$

$\sqrt{3}$ $\displaystyle \sqrt{3}$

$\frac{\sqrt{3}}{9}$ $\displaystyle \frac{\sqrt{3}}{9}$

$3\sqrt{3}$ $\displaystyle 3\sqrt{3}$

Correct answer:

$3\sqrt{3}$ $\displaystyle 3\sqrt{3}$

Explanation:

$a(t)=\frac{d}{dt}v(t)$ $\displaystyle a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

Taking the derivative of the function

$\displaystyle v(t)=9ln(sin(t))$

We'll need to make use of the following derivative rule(s):

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$ $\displaystyle d[ln(u)]=\frac{du}{u}$

Trigonometric derivative:

$\displaystyle d[sin(u)]=cos(u)du$

Note that u may represent large functions, and not just individual variables!

The acceleration function is

$a(t)=9\frac{cos(t)}{sin(t)}=9cot(t)$ $\displaystyle a(t)=9\frac{cos(t)}{sin(t)}=9cot(t)$

At time $t=\frac{\pi}{3}$ $\displaystyle t=\frac{\pi}{3}$

$a(\frac{\pi}{3})=9cot(\frac{\pi}{3})=3\sqrt{3}$ $\displaystyle a(\frac{\pi}{3})=9cot(\frac{\pi}{3})=3\sqrt{3}$

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Example Question #673 : Spatial Calculus

The position of a particle is given by the function $\displaystyle p(t)=(t^3+5)^3$. What is the acceleration of the particle at time t=1 $\displaystyle t=1$ ?

Possible Answers:

1134 $\displaystyle 1134$

540 $\displaystyle 540$

1296 $\displaystyle 1296$

1080 $\displaystyle 1080$

648 $\displaystyle 648$

Correct answer:

1296 $\displaystyle 1296$

Explanation:

$a(t)=\frac{d}{dt}v(t)$ $\displaystyle a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

Taking the derivative of the function

$\displaystyle p(t)=(t^3+5)^3$

We'll need to make use of the Product rule:

$\displaystyle d[uv]=udv+vdu$

Note that u and v may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$\displaystyle v(t)=3(3t^2)(t^3+5)^2$

The acceleration function is

$\displaystyle a(t)=3(9t)(t^3+5)^2+6(3t^2)^2(t^3+5)$

$\displaystyle a(t)=27t(t^3+5)^2+54t^4(t^3+5)$

At time t=1 $\displaystyle t=1$

$\displaystyle a(1)=27(1)(1^3+5)^2+54(1)^4(1^3+5)=1296$

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Example Question #674 : Spatial Calculus

The position of a particle is given by the function $\displaystyle p(t)=-4ln(cos(t))$. What is the acceleration of the particle at time $t=\pi$ $\displaystyle t=\pi$ ?

Possible Answers:

$-\pi$ $\displaystyle -\pi$

$\displaystyle 0$

$\displaystyle 4$

$\displaystyle -4$

$\pi$ $\displaystyle \pi$

Correct answer:

$\displaystyle 4$

Explanation:

$a(t)=\frac{d}{dt}v(t)$ $\displaystyle a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

Taking the derivative of the function

$\displaystyle p(t)=-4ln(cos(t))$

We'll need to make use of the following derivative rule(s):

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$ $\displaystyle d[ln(u)]=\frac{du}{u}$

Trigonometric derivative:

$\displaystyle d[cos(u)]=-sin(u)du$

$d[tan(u)]=\frac{du}{cos^2(u)}$ $\displaystyle d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=-4\frac{-sin(t)}{cos(t)}=4tan(t)$ $\displaystyle v(t)=-4\frac{-sin(t)}{cos(t)}=4tan(t)$

The acceleration function is

$a(t)=\frac{4}{cos^2(t)}$ $\displaystyle a(t)=\frac{4}{cos^2(t)}$

At time $t=\pi$ $\displaystyle t=\pi$

$a(\pi)=\frac{4}{cos^2(\pi)}=4$ $\displaystyle a(\pi)=\frac{4}{cos^2(\pi)}=4$

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Example Question #281 : How To Find Acceleration

The position of a particle is given by the function $\displaystyle p(t)=12t^2+6$. What is the acceleration of the particle at time t=2 $\displaystyle t=2$ ?

Possible Answers:

$\displaystyle 12$

$\displaystyle 24$

$\displaystyle 28$

$\displaystyle 26$

$\displaystyle 20$

Correct answer:

$\displaystyle 24$

Explanation:

$a(t)=\frac{d}{dt}v(t)$ $\displaystyle a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

Taking the derivative of the function

$\displaystyle p(t)=12t^2+6$

The velocity function is

$\displaystyle v(t)=24t$

The acceleration function is

a(t)=24 $\displaystyle a(t)=24$

At time t=2 $\displaystyle t=2$ and at all other times

a=24 $\displaystyle a=24$

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Example Question #676 : Spatial Calculus

The position of a particle is given by the function $\displaystyle p(t)=5sin^2(3t)$. What is the acceleration of the particle at time $t=2\pi$ $\displaystyle t=2\pi$ ?

Possible Answers:

$\displaystyle 45$

$\displaystyle 90$

$\displaystyle -90$

$\displaystyle 0$

$\displaystyle -45$

Correct answer:

$\displaystyle 90$

Explanation:

$a(t)=\frac{d}{dt}v(t)$ $\displaystyle a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$ $\displaystyle v(t)=\frac{d}{dt}p(t)$

Taking the derivative of the function

$\displaystyle p(t)=5sin^2(3t)$

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

$\displaystyle d[sin(u)]=cos(u)du$

$\displaystyle d[cos(u)]=-sin(u)du$

Product rule: $\displaystyle d[uv]=udv+vdu$

Note that u and v may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$\displaystyle v(t)=30sin(3t)cos(3t)$

The acceleration function is

$\displaystyle a(t)=90cos^2(3t)-90sin^2(3t)$

At time $t=2\pi$ $\displaystyle t=2\pi$

$a(2\pi)=90cos^2(3(2\pi))-90sin^2(3(2\pi))=90$ $\displaystyle a(2\pi)=90cos^2(3(2\pi))-90sin^2(3(2\pi))=90$

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Example Question #281 : How To Find Acceleration

The position of a certain point is given by the following function:

$p(t)= \frac{1}{3}t^3 + ln(t) - 23$ $\displaystyle p(t)= \frac{1}{3}t^3 + ln(t) - 23$

What is the acceleration at t=3 $\displaystyle t=3$?

Possible Answers:

$\displaystyle 0$

1.09 $\displaystyle 1.09$

2.5 $\displaystyle 2.5$

5.9 $\displaystyle 5.9$

6.4 $\displaystyle 6.4$

Correct answer:

5.9 $\displaystyle 5.9$

Explanation:

In order to find the acceleration of a certain point, you must first find the derivative of the position function which gives us the velocity function and then the derivative of the velocity function which gives us the acceleration function: $\displaystyle p''(t)= v'(t)= a(t)$

In this case, the position function is: $p(t)= \frac{1}{3}t^3 + ln(t) - 23t$ $\displaystyle p(t)= \frac{1}{3}t^3 + ln(t) - 23t$

The velocity function is found by taking the derivative of the position function: $v(t)= t^2 + \frac{1}{t} - 23$ $\displaystyle v(t)= t^2 + \frac{1}{t} - 23$

The acceleration function is found by taking the derivative of the velocity function: $a(t)= 2t - t^{(-2)}$ $\displaystyle a(t)= 2t - t^{(-2)}$

Finally, to find the accelaration, substitute t=3 $\displaystyle t=3$ into the acceleration function: $a(t)= 2(3) - (3)^{(-2)}$ $\displaystyle a(t)= 2(3) - (3)^{(-2)}$

Therefore, the answer is: 5.9 $\displaystyle 5.9$

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Example Question #281 : Acceleration

The position of a certain point is given by the following function:

$p(t)= 6t^{-\frac{1}{6}} + 4t^5 - 4t^2$ $\displaystyle p(t)= 6t^{-\frac{1}{6}} + 4t^5 - 4t^2$

What is the acceleration at t=1 $\displaystyle t=1$?

Possible Answers:

73.2 $\displaystyle 73.2$

64.8 $\displaystyle 64.8$

$\displaystyle 23$

5.9 $\displaystyle 5.9$

$\displaystyle 95$

Correct answer:

73.2 $\displaystyle 73.2$

Explanation:

In this case, the position function is: $p(t)= 6t^{-\frac{1}{6}} + 4t^5 - 4t^2$ $\displaystyle p(t)= 6t^{-\frac{1}{6}} + 4t^5 - 4t^2$

The velocity function is found by taking the derivative of the position function: $v(t)= -t^{-\frac{7}{6}} + 20t^4 - 8t$ $\displaystyle v(t)= -t^{-\frac{7}{6}} + 20t^4 - 8t$

The acceleration function is found by taking the derivative of the velocity function: $a(t)=\frac{7}{6}t^{-\frac{13}{6}} + 80t^3 - 8$ $\displaystyle a(t)=\frac{7}{6}t^{-\frac{13}{6}} + 80t^3 - 8$

Finally, to find the accelaration, substitute t=1 $\displaystyle t=1$ into the acceleration function: $a(1)=\frac{7}{6}(1)^{-\frac{13}{6}} + 80(1)^3 - 8$ $\displaystyle a(1)=\frac{7}{6}(1)^{-\frac{13}{6}} + 80(1)^3 - 8$

Therefore, the answer is: 73.2 $\displaystyle 73.2$

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Example Question #281 : How To Find Acceleration

The position of a certain point is given by the following function:

$p(t)= -\cos t + 3t^2 + 6ln(t)$ $\displaystyle p(t)= -\cos t + 3t^2 + 6ln(t)$

What is the acceleration at $t=\pi$ $\displaystyle t=\pi$?

Possible Answers:

45.3 $\displaystyle 45.3$

73.2 $\displaystyle 73.2$

4.4 $\displaystyle 4.4$

3.08 $\displaystyle 3.08$

100.3 $\displaystyle 100.3$

Correct answer:

4.4 $\displaystyle 4.4$

Explanation:

In this case, the position function is: $p(t)= -\cos t + 3t^2 + 6ln(t)$ $\displaystyle p(t)= -\cos t + 3t^2 + 6ln(t)$

The velocity function is found by taking the derivative of the position function: $v(t)= \sin t + 6t + \frac{6}{t}$ $\displaystyle v(t)= \sin t + 6t + \frac{6}{t}$

The acceleration function is found by taking the derivative of the velocity function: $a(t)= \cos t + 6 - 6t^{-2}$ $\displaystyle a(t)= \cos t + 6 - 6t^{-2}$

Finally, to find the accelaration, substitute $t=\pi$ $\displaystyle t=\pi$ into the acceleration function: $a(\pi)= \cos (\pi) + 6 - 6(\pi)^{-2}$ $\displaystyle a(\pi)= \cos (\pi) + 6 - 6(\pi)^{-2}$

Therefore, the answer is: 4.4 $\displaystyle 4.4$

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Example Question #671 : Spatial Calculus

The position of a certain point is given by the following function:

$p(t)= \frac{2}{3} t^3 + ln(t^2)+ 5$ $\displaystyle p(t)= \frac{2}{3} t^3 + ln(t^2)+ 5$

What is the acceleration at t=1 $\displaystyle t=1$?

Possible Answers:

32.6 $\displaystyle 32.6$

4.4 $\displaystyle 4.4$

$\displaystyle 0$

12.2 $\displaystyle 12.2$

$\displaystyle 2$

Correct answer:

$\displaystyle 2$

Explanation:

In this case, the position function is: $p(t)= \frac{2}{3} t^3 + ln(t^2)+ 5$ $\displaystyle p(t)= \frac{2}{3} t^3 + ln(t^2)+ 5$

The velocity function is found by taking the derivative of the position function: $v(t)= 2 t^2 + \frac{2}{t}$ $\displaystyle v(t)= 2 t^2 + \frac{2}{t}$

The acceleration function is found by taking the derivative of the velocity function: $a(t)= 4 t - 2t^{-2}$ $\displaystyle a(t)= 4 t - 2t^{-2}$

Finally, to find the accelaration, substitute t=1 $\displaystyle t=1$ into the acceleration function: $a(1)= 4 (1) - 2(1)^{-2}$ $\displaystyle a(1)= 4 (1) - 2(1)^{-2}$

Therefore, the answer is: $\displaystyle 2$

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #280 : How To Find Acceleration

Example Question #672 : Spatial Calculus

Example Question #673 : Spatial Calculus

Example Question #674 : Spatial Calculus

Example Question #281 : How To Find Acceleration

Example Question #676 : Spatial Calculus

Example Question #281 : How To Find Acceleration

Example Question #281 : Acceleration

Example Question #281 : How To Find Acceleration

Example Question #671 : Spatial Calculus

All Calculus 1 Resources

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