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Calculus 1 : How to find acceleration

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Example Questions

Example Question #171 : Acceleration

Find the acceleration function of a particle if its velocity function is v(t)=7t-7cos(t) .

Possible Answers:

a(t)=7t

a(t)=7

a(t)=7+7sin(t)

a(t)=-7cos(t)

a(t)=7sin(t)

Correct answer:

a(t)=7+7sin(t)

Explanation:

The acceleration function is the first derivative of the velocity function, or

$a(t)=v'(t)=\frac{d}{dt}(v(t))$ .

We will use the power rule, $\frac{d}{dt}(t^{c})=ct^{c-1}$ , where is a constant,

and one of the trigonometric rules, $\frac{d}{dt}(cos(t))=-sin(t)$ , to find the derivative of this velocity function.

For this problem:

$a(t)=\frac{d}{dt}(v(t))=7\frac{d}{dt}(t)-7\frac{d}{dt}(cos(t))$

a(t)=7+7sin(t)

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Example Question #171 : How To Find Acceleration

Find the acceleration function of a car if its position function is $s(t)=t^{3}-t^{2}-15t-8$ .

Possible Answers:

$a(t)=3t^{2}$

a(t)=6t-2

a(t)=15

$a(t)=3t^{2}-2t-15$

$a(t)=t^{3}-t^{2}$

Correct answer:

a(t)=6t-2

Explanation:

The acceleration function is the second derivative of the position function, or

$a(t)=s''(t)=\frac{d^{2}}{dt^{2}}(s(t))=\frac{d}{dt}\left [ \frac{d}{dt}(s(t)) \right ]$ .

We will use the power rule, $\frac{d}{dt}(t^{c})=ct^{c-1}$ , where is a constant,

and the constant rule, $\frac{d}{dt}(c)=0$ , where is a constant, to find the first and second derivative of this position function.

For this problem:

$s(t)=t^{3}-t^{2}-15t-8$

$\frac{d}{dt}(s(t))=\frac{d}{dt}(t^{3})-\frac{d}{dt}(t^{2})-15\frac{d}{dt}(t)-\frac{d}{dt}(8)$

$\frac{d}{dt}(s(t))=3t^{2}-2t-15$

$a(t)=\frac{d}{dt}\left [ \frac{d}{dt}(s(t)) \right ]=3\frac{d}{dt}(t^{2})-2\frac{d}{dt}(t)-\frac{d}{dt}(15)$

a(t)=3*2t-2

a(t)=6t-2

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Example Question #173 : How To Find Acceleration

Find the acceleration function of a particle if its position function is $s(t)=t^{3}+2sin(3t)$ .

Possible Answers:

a(t)=6t-18sin(3t)

a(t)=18

a(t)=18sin(3t)

$a(t)=t^{3}$

a(t)=2sin(3t)

Correct answer:

a(t)=6t-18sin(3t)

Explanation:

The acceleration function is the second derivative of the position function, or

$a(t)=s''(t)=\frac{d^{2}}{dt^{2}}(s(t))=\frac{d}{dt}\left [ \frac{d}{dt}(s(t)) \right ]$ .

We will use the power rule, $\frac{d}{dt}(t^{c})=ct^{c-1}$ , where is a constant,

and the trigonometric rules, $\frac{d}{dt}(cos(at))=-asin(at)$ and $\frac{d}{dt}(sin(at))=acos(at)$ , where is a constant, to find the first and second derivative of this position function.

For this problem:

$s(t)=t^{3}+2sin(3t)$

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

$\frac{d}{dt}(s(t))=3t^{2}+2*[3cos(3t)]$

$\frac{d}{dt}(s(t))=3t^{2}+6cos(3t)$

$a(t)=\frac{d}{dt}\left [ \frac{d}{dt}(s(t)) \right ]=3\frac{d}{dt}(t^{2})+6\frac{d}{dt}[cos(3t)]$

a(t)=3*2t+6[-3sin(3t)]

a(t)=6t-18sin(3t)

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Example Question #173 : How To Find Acceleration

$x(t)=\frac{1}{t}$

A particle's position is represented by the above function, x(t) . Find the function representing the particle's acceleration.

Possible Answers:

$a(t)=\frac{-1}{t^3}$

a(t)=-ln(t)

a(t)=ln(t)

$a(t)=\frac{2}{t^3}$

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

Correct answer:

$a(t)=\frac{2}{t^3}$

Explanation:

To find the function representing acceleration, differentiate the position function twice. Use the power rule to differentiate.

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

The first derivative is taken as follows:

$x(t)=\frac{1}{t}$

$x'(t)=-\frac{1}{t^2}$

Then, differentate x'(t) to get the second derivative.

$x''(t)=\frac{2}{t^3}$

Therefore, the acceleration function is:

$a(t)=\frac{2}{t^3}$

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Example Question #174 : How To Find Acceleration

v(t)=4sin(t)

The velocity of a pendulum is given by the velocity function above, v(t) .

Find the pendulum's acceleration when $t=2\pi$ .

Possible Answers:

Correct answer:

Explanation:

To find the funciton representing the pendulum's acceleration, differentiate the velocity function.

v(t)=4sin(t)

v'(t)=4sin(t)

Therefore, the acceleration function is:

a(t)=4cos(t)

Evaluate a(t) when $t=2\pi$ . In other words, evaluate $a(2\pi)$ .

$a(2\pi)=4cos(2\pi)$

$a(2\pi)=4*1$

$a(2\pi)=4$

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Example Question #171 : How To Find Acceleration

The position of a particle as a function of time is s(t)=2t^3+t^4 . What is the acceleration of the particle when t=1 ?

Possible Answers:

$\tfrac{11}{12}$

Correct answer:

Explanation:

The acceleration is the second derivative of the position and the derivative of the velocity, which in turn is the derivative of the position.

$\\ a(t)=v'(t)=s''(t)\\ \\=(2t^3+t^4)''=(6t^2+4t^3)'\\ \\=12t+12t^2$

At t=1 , we have

a(1)=12(1)+12(1)^2=24 .

Note: these derivatives can be evaluated using the Power Rule --- which says that for

p(x)=x^n , $p'(x)=nx^{n-1}$ --- and the linear properties of derivatives.

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Example Question #173 : How To Find Acceleration

Find the acceleration at t=0 of the particle if its velocity is given by the following function:

$v(t)=10\cos(t)+te^t+2t^2$

Possible Answers:

Correct answer:

Explanation:

The acceleration of the particle is given by the first derivative of the velocity function, which is

$v'(t)=a(t)=-10\sin(t)+e^t+te^t+4t$

and was found using the following rules:

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

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

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

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

Now, plug in t=0 into the above function and we get

a(0)=0+1+0+0=1

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Example Question #171 : How To Find Acceleration

A particle moves in space with the following position function:

s(t)=e^t+t^3+4t

What is the acceleration at t=1 ?

Possible Answers:

e^1+6

e^1+5

e^1+7

Correct answer:

e^1+6

Explanation:

The acceleration function is the second derivative of the position function, or the first derivative of the velocity function, which itself is the first derivative of the position function. So, we must first find the velocity function, which is

v(t)=s'(t)=e^t+3t^2+4

and was found using the following rules:

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

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

Now, take the derivative of the velocity function to find the acceleration:

v'(t)=a(t)=e^t+6t

and was found using the same rules as above.

Finally, plug in t=1 into the above function:

a(1)=e^1+6

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Example Question #570 : Calculus

The position of a particle moving in a straight horizontal line is given by the equation

$x(t)=\ln (\frac{t^2+1}{e^{2t}})$ .

For what value is the acceleration of the particle zero? Particle starts to move at t=0 .

Possible Answers:

t=1

t=3

$t=\frac{1}{2}$

$t=\frac{3}{2}$

t=2

Correct answer:

t=1

Explanation:

To find the acceleration of the particle, find the second derivative of the position equation.

For this particular function we will need to use the power rule, natural log rule, quotient rule, and exponential rule.

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

Natural Log Rule: $ln(u)\rightarrow \frac{1}{u}\cdot \frac{du}{dx}$

Quotient Rule: $\frac{f(x)}{g(x)}\rightarrow \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

Exponential Rule: $e^u\rightarrow e^u \cdot \frac{du}{dx}$

First lets use the rule of logs to rewrite the position function as follows.

$x(t)=\ln (\frac{t^2+1}{e^{2t}})=\ln (t^2+1)-2t$ .

Now applying the above rules of differentiation the first derivative will give you the velocity of the particle.

$x'(t)=v(t)=\frac{2t}{t^2+1}-2$ .

The derivative of the velocity will then give you the acceleration of the particle.

$v'(t)=a(t)=\frac{2(1-t^2)}{(t^2+1)^{2}}$ .

Then set the accelartion equal to zero and solve for the time. Only choose the time that is applicable $(t\geq0)$ .

$a(t)=\frac{2(1-t^2)}{(t^2+1)}=0$ .

And when you solve for it, t=1 .

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Example Question #171 : 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:

$25\: m/s^2$

$6\: m/s^2$

$16\: m/s^2$

$3\: m/s^2$

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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Calculus 1 : How to find acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #171 : Acceleration

Example Question #171 : How To Find Acceleration

Example Question #173 : How To Find Acceleration

Example Question #173 : How To Find Acceleration

Example Question #174 : How To Find Acceleration

Example Question #171 : How To Find Acceleration

Example Question #173 : How To Find Acceleration

Example Question #171 : How To Find Acceleration

Example Question #570 : Calculus

Example Question #171 : Acceleration

All Calculus 1 Resources

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