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Calculus 1 : Acceleration

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Example Questions

Example Question #571 : Spatial Calculus

Given an object moving with velocity

$v(t)=5sin(e^{2t})+2ln(t)$ .

Find the acceleration at t=2 .

Possible Answers:

$-10e^{4}cos(e^{4})+1$

$10cos(e^{4})+1$

$5cos(e^{4})+1$

$10e^{4}cos(e^{4})+1$

Correct answer:

$10e^{4}cos(e^{4})+1$

Explanation:

Acceleration a(t) is the derivative of velocity v(t) with respect to time.

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

In our case, $v(t)=5sin(e^{2t})+2ln(t)$

By the sum rule,

$\frac{\mathrm{d} }{\mathrm{d} t}(5sin(e^{2t})+2ln(t))=\frac{\mathrm{d} }{\mathrm{d} t}(5sin(e^{2t})+ \frac{\mathrm{d} }{\mathrm{d} t}(2ln(t))$

By the chain rule, we know that

$\frac{d}{dt}f(g(t))=f'(g(t))*g'(t)$

For ,

$\frac{\mathrm{d} }{\mathrm{d} t}(5sin(e^{2t})$ ,

f(t)=5sin(t) ,

$g(t)=e^{2t}$ ,

f'(t)=5cos(t) ,

$g'(t)=2e^{2t}$

Therefore,

$\frac{\mathrm{d} }{\mathrm{d} t}(5sin(e^{2t})= 10e^{2t}cos(e^{2t})$

For $\frac{\mathrm{d} }{\mathrm{d} t}(2ln(t))$ , we know that this is equal to

$2\frac{\mathrm{d} }{\mathrm{d} t}(ln(t))=\frac{2}{t}$

Therefore, the acceleration is given by

$a(t)=10e^{2t}cos(e^{2t})+\frac{2}{t}$

Since we need the acceleration at

$a(2)=10e^{4}cos(e^{4})+1$

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Example Question #181 : Acceleration

Find the acceleration function a(t) if

v(t)=3t^2+6t .

Possible Answers:

a(t)=2t+3

a(t)=3t+6

a(t)=3t+6t

a(t)=6t+6

Correct answer:

a(t)=6t+6

Explanation:

In order to find the acceleration function from the velocity function we need to find the derivative of the velocity function since

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

When taking the derivative, we will use the power rule which states

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

and by applying this rule to each term we get

$\frac{d}{dt}v(t)=\frac{d}{dt}[3t^2+6t]$

$=2*3t^{2-1}+1*6t^{1-1}$

=6t+6 .

Hence,

a(t)=6t+6 .

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Example Question #572 : Calculus

Find the acceleration function if

$v(t)=tan \,t$ .

Possible Answers:

a(t)=sec^2 t

$a(t)=cos\, t$

$a(t)=csc\, t$

$a(t)=sin\, t$

Correct answer:

a(t)=sec^2 t

Explanation:

In order to find the acceleration function from the velocity function we need to find the derivative of the velocity function since

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

When taking the derivative, we will use the trigonometric derivative which states

$\frac{\mathrm{d} }{\mathrm{d} x}tan\,x=sec^2x$

and by applying this rule we get

$\frac{d}{dt}v(t)=\frac{d}{dt}[tan\, t]$

=sec^2t .

Hence,

a(t)=sec^2 t .

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Example Question #573 : Calculus

Find the acceleration function a(t) if

v(t)=10t .

Possible Answers:

a(t)=1

a(t)=10

a(t)=t

a(t)=5t^2

Correct answer:

a(t)=10

Explanation:

In order to find the acceleration function from the velocity function we need to find the derivative of the velocity function since

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

When taking the derivative, we will use the power rule which states

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

and by applying this rule to each term we get

$\frac{d}{dt}v(t)=\frac{d}{dt}[10t]$

$=10t^{1-1}$ .

Hence,

a(t)=10 .

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Example Question #574 : Calculus

What is the acceleration of Pym the particle at time t=0 if his position function is x(t)=3t^2+sin(t^2) ?

Possible Answers:

Correct answer:

Explanation:

Acceleration is the second time derivative of position:

$a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)$

For the position function use the power rule which states,

$x^n \ dx\rightarrow nx^{n-1}$

and the rule for trigonometric functions which state that the derivative of sine is cosine and the derivative of cosine is a negative sine.

Applying these rules twice to our function,

x(t)=3t^2+sin(t^2)

we can find the acceleration.

The velocity and acceleration functions are:

v(t)=6t+2tcos(t^2)

a(t)=6+2cos(t^2)-4t^2sin(t^2)

a(0)=6+2cos(0^2)-4(0^2)sin(0^2)

a(0)=8

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Example Question #186 : How To Find Acceleration

The position of a particle is given by the functions x(t)=2t+3 , y(t)=2x^2(t) .

What is the acceleration of the particle, in Cartesian coordinates, at time t=4 ?

Possible Answers:

(0,16)

(4,32)

(0,0)

(4,4)

(4,16)

Correct answer:

(0,16)

Explanation:

Acceleration is the second time derivative of position, or the first time derivative of velocity:

$a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}p(t)$

For the position functions:

x(t)=2t+3

y(t)=2x^2(t)

The velocity functions are:

v_x(t)=2

$v_y(t)=4x(t)\frac{dx(t)}{dt}$

And the acceleration functions are:

a_x(t)=0

$a_y(t)=4\frac{dx(t)}{dt}\frac{dx(t)}{dt}+4x(t)\frac{d^2x(t)}{dt^2}$

Now, note that $\frac{dx(t)}{dt}=v_x(t)$ and $\frac{d^2x(t)}{dt^2}=a_x(t)$ . This allows simplification of the a_y(t) function:

a_y(t)=4(2)(2)+4(2t+3)(0)

a_y(t)=16

As it turns out, both acceleration functions are constant!

a_x(4)=0

a_y(4)=16

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Example Question #575 : Calculus

The position of a particle in two dimensions is given by the functions x(t)=3t^3-5t^2 , $y(t)=e^{-2t}cos(t)$ . What is the particle's acceleration?

Possible Answers:

$(18t,3e^{-2t}cos(t))$

$(18t-10,-4e^{-2t}sin(t)+3e^{-2t}cos(t))$

$(18t,4e^{-2t}sin(t))$

$(-10,4e^{-2t}sin(t))$

$(18t-10,4e^{-2t}sin(t)+3e^{-2t}cos(t))$

Correct answer:

$(18t-10,4e^{-2t}sin(t)+3e^{-2t}cos(t))$

Explanation:

Acceleration is the first time derivative of velocity and the second time derivative of position:

$a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}p(t)$

For the position functions

x(t)=3t^3-5t^2

$y(t)=e^{-2t}cos(t)$

The velocity functions are:

v_x(t)=9t^2-10t

$v_y(t)=-2e^{-2t}cos(t)-e^{-2t}sin(t)$

And therefore the acceleration functions are:

a_x(t)=18t-10

$a_y(t)=4e^{-2t}cos(t)+2e^{-2t}sin(t)+2e^{-2t}sin(t)-e^{-2t}cos(t)$

$a_y(t)=4e^{-2t}sin(t)+3e^{-2t}cos(t)$

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Example Question #572 : Spatial Calculus

The position of a particle is given by the function x(t)=t^3+tcos(2t) . What is its acceleration at time $t=\pi$ ?

Possible Answers:

$2\pi+8$

$2\pi$

$6\pi+4$

$6\pi$

$6\pi-4$

Correct answer:

$2\pi$

Explanation:

Acceleration is the second time derivative of position, and the first time derivative of velocity:

$a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)$

To find the acceleration function we will need to use the power rule, chain rule, and trigonometric rules.

Power Rule: $x^n \ dx =nx^{n-1}$

Chain Rule: $f(g(x))=f'(g(x))\cdot g'(x)$

Trigonometric Rule for cosine: cos(t)dt=-sin(t)

Trigonometric Rule for sine: sin(t)dt=cos(t)

Applying the above rules twice we will find our acceleration function.

For position function

x(t)=t^3+tcos(2t)

v(t)=3t^2-2tsin(2t)+cos(2t)

a(t)=6t-4tcos(2t)-2sin(2t)-2sin(2t)

a(t)=6t-4tcos(2t)-4sin(2t)

$a(\pi)=6\pi-4\pi$

$a{(\pi)}=2\pi$

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Example Question #576 : Calculus

The velocity of an automobile is given by:

$f(x)=e^{3x}\cos(x)+2x^2$

What is the acceleration of the automobile?

Possible Answers:

$e^{3x}\cos(x)-e^{3x}\sin(x)+4x$

$3e^{3x}\cos(x)-e^{3x}\sin(x)+2x^2$

$-3e^{3x}\sin(x)+4x$

$3e^{3x}\cos(x)-e^{3x}\sin(x)+4x$

Correct answer:

$3e^{3x}\cos(x)-e^{3x}\sin(x)+4x$

Explanation:

The acceleration of the automobile is given by the derivative of the velocity function:

$f'(x)=3e^{3x}\cos(x)+e^{3x}\sin(x)+4x$

We used the following rules to find the derivative:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\rightarrow (2x^2)'=2\cdot 2x^{2-1}=4x$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{du}{dx}$ .

In this particular case let

$f(x)=e^{3x}\rightarrow f'(x)=3e^3x$

and

$g(x)=cos(x)\rightarrow g'(x)=-sin(x)$ .

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Example Question #181 : Acceleration

What is the acceleration funtion of a particle if the position of the particle is represented by the function:

$\int_{x}^{6}({x}^2+5x-6)dx$

Possible Answers:

x^2+5x-6

2x+5

-2x-5

Correct answer:

-2x-5

Explanation:

The acceleration function is the sencond derivative of the position function, so we must derive the given equation twice.

One of the properties of integrals is $\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$ ,

$\int_{x}^{6}({x}^2+5x-6)dx=-\int_{6}^{x}({x}^2+5x-6)dx$ .

By the fundemental theorem of Calculus, the derivative of $\int_{a}^{x}f(x)dx$ is f(x) , so

$\left(-\int_{6}^{x}({x}^2+5x-6)dx\right)'=-\left(\int_{6}^{x}({x}^2+5x-6)dx\right)'=-({x}^2+5x-6)$

Then, the second derivative, using the power rule, $\frac{d}{dx}x^n=nx^{n-1}$ is $-({x}^2+5x-6)'=-(2x+5)=-2x-5$ .

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #571 : Spatial Calculus

Example Question #181 : Acceleration

Example Question #572 : Calculus

Example Question #573 : Calculus

Example Question #574 : Calculus

Example Question #186 : How To Find Acceleration

Example Question #575 : Calculus

Example Question #572 : Spatial Calculus

Example Question #576 : Calculus

Example Question #181 : Acceleration

All Calculus 1 Resources

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