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Calculus 1 : Calculus

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Example Questions

Example Question #172 : How To Find Acceleration

A rock is rolling down a steep hill. If the distance (in meters) it has traveled in seconds can be described by the function s(t) = 3t^2+1 , how fast is it accelerating if it has been rolling for five seconds?

Possible Answers:

$3\: m/s^2$

$16\: m/s^2$

$6\: m/s^2$

$25\: m/s^2$

Correct answer:

$6\: m/s^2$

Explanation:

If we consider the relation between acceleration and position, we can see that it is the rate of change of the rate of change of the position.

So this problem amounts to calculating the second derivative of our function at the given point.

$(x^n)' = n\cdot x^{n-1}$ and the derivative of a constant is 0, so:

s(t) = 3t^2+1

s'(t)=6t

s''(t)=6

Fortunately for us this is a constant function, and so no more calculation is needed. 6 itself is the answer.

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Example Question #181 : How To Find Acceleration

Given an object moving with velocity

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

Find the acceleration at t=2 .

Possible Answers:

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

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

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

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

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)=3t+6t

a(t)=2t+3

a(t)=6t+6

a(t)=3t+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 #182 : How To Find Acceleration

Find the acceleration function if

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

Possible Answers:

$a(t)=csc\, t$

$a(t)=sin\, t$

$a(t)=cos\, t$

a(t)=sec^2 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 #183 : Acceleration

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 #184 : Acceleration

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 #185 : Acceleration

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:

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

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

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

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

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

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 #182 : How To Find Acceleration

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:

$6\pi-4$

$6\pi$

$2\pi$

$6\pi+4$

$2\pi+8$

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 #186 : Acceleration

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:

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

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

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

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

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #172 : How To Find Acceleration

Example Question #181 : How To Find Acceleration

Example Question #181 : Acceleration

Example Question #182 : How To Find Acceleration

Example Question #183 : Acceleration

Example Question #184 : Acceleration

Example Question #186 : How To Find Acceleration

Example Question #185 : Acceleration

Example Question #182 : How To Find Acceleration

Example Question #186 : Acceleration

All Calculus 1 Resources

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