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Calculus 1 : Acceleration

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Example Questions

Example Question #121 : Acceleration

A snowball rolls into a valley. Its position is represented by s(t)=-16t^2+36t+12 , where represents distance in terms of meters and represents time in terms of seconds.

What is the acceleration of the snowball?

Possible Answers:

$-32\frac{m}{s^2}$

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

$32\frac{m}{s^2}$

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

Correct answer:

$-32\frac{m}{s^2}$

Explanation:

Take the second derivative, using the Power Rule ( $x^n=nx^{n-1}$ ) twice, of s(t) :

s'(t)=-32t+36

s''(t)=-32

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Example Question #122 : Acceleration

A weight hanging from a spring is stretched down 3 units beyond its rest position and released at time t=0 to bob up and down. Its position at any later time is

s=3cos(t)

What is the acceleration at time ?

Possible Answers:

-3sin(t)

3sin(t)

3cos(t)

-3cos(t)

sin(t)

Correct answer:

-3cos(t)

Explanation:

$v=\frac{ds}{dt}=\frac{d}{dt}(3cos(t))=-3sin(t)$

$a = \frac{dv}{dt}=\frac{d}{dt}(-3sin(t))=-3cos(t)$

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Example Question #123 : Acceleration

The position of a particle is given by the expression

$x(t)=t^{3}+3t+3$ .

At what time is the acceleration of the particle ?

Possible Answers:

t=5

t=3

t=-3

t=7

Correct answer:

t=5

Explanation:

To find the acceleration of a particle given the equation for position x(t), utilize the relationship

a(t)=v'(t)=x''(t)

Therefore, to find the acceleration, we take the second derivative of x(t):

$v(t)=x'(t)=3t^{2}+3$

a(t)=v'(t)=x''(t)=6t

Now to find where the acceleration is 30, we set 6t equal to 30 and solve for t:

6t=30

t=5

Which gives us the correct answer.

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Example Question #124 : Acceleration

The velocity of a particle is given by the equation

$v(t)=e^{\sin (5t)}$ .

To two decimal places, what is the acceleration of the particle at t=3 ?

Possible Answers:

4.34

-7.28

-1.52

-1.14

Correct answer:

-7.28

Explanation:

The acceleration of a particle is given by the first derivative of the velocity:

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

Here, the chain rule must be used to fine the derivative of the velocity:

$\frac{dv(t)}{dt}=e^{\sin (5t)}\cdot \cos(5t)\cdot 5=a(t)$

Finally, evaluate the expression for acceleration at t=3:

$a(3)=e^{\sin(5\cdot3)}\cdot\cos(5\cdot3)\cdot5=-7.28$

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Example Question #125 : Acceleration

Find f''(t) at t=0 , where

$f(t)=te^{t^3+1}$ .

Possible Answers:

Correct answer:

Explanation:

In order to evaluate at t=0 , we need to find f''(t) .

$f(t)=te^{t^3+1}$

$f'(t)=e^{t^3+1}+3t^3e^{t^3+1}$

We found this by using the definition of derivatives for exponential functions and the product rule.

Remember that the definition of derivaties for exponential functions are

$\\ g(t)=e^{h(t)} \\ g'(t)=h'(t)e^{h(t)}$ .

Remember that the product rule is as follows:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

Now lets find f''(t) .

$f'(t)=e^{t^3+1}+3t^3e^{t^3+1}$

$f''(t)=3t^2e^{t^3+1}+9t^2e^{t^3+1}+9t^5e^{t^3+1}$

Now lets evaluate f''(0) .

$\\ f''(0)=3(0)^2e^{0^3+1}+9(0)^2e^{0^3+1}+9(0)^5e^{0^3+1} \\ \\ f''(0)=0+0+0=0$

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Example Question #126 : Acceleration

Find the acceleration of a particle at t = 0 with a position defined by $s(t) = t^3 - e^{\cos(3t)} + 17$ .

Possible Answers:

9e + 6

3e+6

Correct answer:

Explanation:

Acceleration of a particle is given by the second derivative of the position function. The position function is

$s(t) = t^3 - e^{\cos(3t)} + 17$

Taking the first derivative gives the velocity function:

$s'(t) = v(t) = 3t^2 + 3\sin(3t)e^{\cos (3t)}$

Taking the second derivative gives the acceleration function:

$s''(t) = v'(t) = a(t) = 6t- 9\sin ^{2}(3t)e^{\cos (3t)} + 9\cos(3t)e^{cos(3t)}$

Evaluating the function at t = 0 gives the acceleration at this time:

$s''(0) = a(0) = 6(0) - 9\sin^{2}(0)e^{\cos{0}} + 9cos(0)e^{\cos{0}} = 9e$

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Example Question #127 : Acceleration

A given ball has a position defined by the equation $p(t)=4t^{3}-5t^{2}+3t+10$ . What is its acceleration at time t=2 ?

Possible Answers:

Correct answer:

Explanation:

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

According to the Power Rule, the first derivative

$\frac{d}{dt}t^{n}=nt^{n-1}$ for all $n\neq0$ ;

therefore, the second derivative

$\frac{d}{dt}nt^{n-1}=(n-1)nt^{n-2}$ , $n\neq0$ .

Applying this to

$p(t)=4t^{3}-5t^{2}+3t+10$ , we get

$p'(t)=12t^{2}-10t+3$ and

a(t)=p''(t)=24t-10 .

Plugging in t=2 ,

a(2)=p''(2)=24(2)-10=48-10=38 .

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Example Question #128 : Acceleration

A given propeller plane has a position defined by the equation $p(t)=2t^{3}+3t^{2}-12t+2$ . What is its acceleration at time t=3 ?

Possible Answers:

Correct answer:

Explanation:

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

According to the Power Rule, the first derivative

$\frac{d}{dt}t^{n}=nt^{n-1}$ for all $n\neq0$ ;

therefore, the second derivative

$\frac{d}{dt}nt^{n-1}=(n-1)nt^{n-2}$ , $n\neq0$ .

Applying this to

$p(t)=2t^{3}+3t^{2}-12t+2$ , we get

$p'(t)=6t^{2}+6t-12$ and

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

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

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Example Question #129 : Acceleration

A given object has a position defined by the equation $p(t)=t^{3}+10t^{2}+5t+7$ . What is its acceleration at time t=5 ?

Possible Answers:

Correct answer:

Explanation:

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

According to the Power Rule, the first derivative

$\frac{d}{dt}t^{n}=nt^{n-1}$ for all $n\neq0$ ;

therefore, the second derivative

$\frac{d}{dt}nt^{n-1}=(n-1)nt^{n-2}$ , $n\neq0$ .

Applying this to

$p(t)=t^{3}+10t^{2}+5t+7$ , we get $p'(t)=3t^{2}+20t+5$ and a(t)=p''(t)=6t+20 .

Plugging in t=5 , a(5)=p''(5)=6(5)+20=30+20=50 .

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Example Question #521 : Spatial Calculus

If the position of particle is $\small s(t)=te^{4t}$ , then what is its acceleration function $\small a(t)$ ?

Possible Answers:

$\small a(t)=(8+16t)e^{4t}$

$\small \small a(t)=16e^{4t}$

$\small \small a(t)=0$

$\small \small a(t)=(8+8t)e^{4t}$

Correct answer:

$\small a(t)=(8+16t)e^{4t}$

Explanation:

Since the acceleration is the second derivative of the position $\small s(t)=te^{4t}$ , we must differentiate it twice, with the help of the product rule:

$\small \small s'(t)=e^{4t}+t\cdot 4e^{4t}=(1+4t)e^{4t}$

$\small \small \small \implies s''(t)=a(t)=4e^{4t}+(1+4t)\cdot 4e^{4t}$

$\small =4e^{4t}+(4+16t)e^{4t}=(8+16t)e^{4t}$

So the acceleration function is $\small a(t)=(8+16t)e^{4t}$ .

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #121 : Acceleration

Example Question #122 : Acceleration

Example Question #123 : Acceleration

Example Question #124 : Acceleration

Example Question #125 : Acceleration

Example Question #126 : Acceleration

Example Question #127 : Acceleration

Example Question #128 : Acceleration

Example Question #129 : Acceleration

Example Question #521 : Spatial Calculus

All Calculus 1 Resources

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