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Calculus 1 : Acceleration

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Example Questions

Example Question #441 : Spatial Calculus

The velocity of an object is given by the equation $v(t) = 4t\cos(t)$ . What is the acceleration of the object at time t = 1 ?

Possible Answers:

$a(1) = 4\cos(1) + 4\sin(1)$

$a(1) = 2\cos(1) + 4\sin(1)$

None of these.

$a(1) = 4\cos(1) - 4\sin(1)$

Correct answer:

$a(1) = 4\cos(1) - 4\sin(1)$

Explanation:

The acceleration of the object can be found by differentiating the velocity equation of the object. To do this we can use the product rule where if

$f(x) = h(x)g(x) \rightarrow f'(x) = h'(x)g(x) + h(x)g'(x)$ .

We must also use the power rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Rewriting the velocity equation to $v(t) = 4t^1\cos(t)$ , we can apply these rules to the velocity equation along with the knowledge that the derivative of $\cos(t)$ is $-\sin(t)$ giving us,

$a(t) = v'(t) = 4(1)t^{1-1}(\cos(t))+4t^1(-\sin(t)) = 4\cos(t)-4t\sin(t)$ .

To find the acceleration at t = 0 we now substitute the value of for in the acceleration equation.

$a(1) = 4\cos(1) - 4(1)\sin(t) = 4\cos(1) - 4\sin(1)$

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Example Question #51 : Acceleration

The velocity of an object is given by the equation $v(t) = \frac{\sin(t)}{t^2}$ . What is the initial acceleration of the object?

Possible Answers:

None of these.

Correct answer:

None of these.

Explanation:

The initial acceleration of the object can be found by differentiating the object's velocity equation $v(t) = \frac{\sin(t)}{t^2}$ and solving for a(0) .

To differentiate the velocity equation, we can use the quotient rule where if

$f(x)= \frac{h(x)}{g(x)} \rightarrow f'(x) = \frac{h'(x)g(x) - h(x)g'(x)}{[g(x)]^2}$ .

Additionally, to find the derivative of the equation we must use the power rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Applying these rules with the knowledge that the derivative of $\sin(t)$ is $\cos(t)$ , we can find the acceleration equation.

$a(t) = v'(t) = \frac{(\cos(t))t^2 - 2t^{2-1}(\sin(t))}{[t^2]^2} = \frac{t^2\cos(t) - 2t\sin(t)}{t^4}$

The acceleration of the object is the value of the acceleration equation when t = 0 .

$a(0) = \frac{0^2\cos(0)-2(0)\sin(0)}{0^4} = \frac{0-0}{0} = \frac{0}{0}$

Therefore the initial acceleration of the object is undefined.

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Example Question #443 : Spatial Calculus

The jerk of an object is $j(t) = \sin(t)$ . What is the acceleration equation of the object if the acceleration is at time t= 4 ?

Possible Answers:

a(t) = 2

$a(t) = -\cos(t)+2+\cos(4)$

$a(t) = 2\cos(t)+2$

$a(t) = \cos(t)+2+\cos(4)$

Correct answer:

$a(t) = -\cos(t)+2+\cos(4)$

Explanation:

The acceleration equation can be found by integrating the jerk equation $j(t) = \sin(t)$ .

The integral of $\sin(t)$ is $-\cos(t)$ .

Therefore the acceleration equation is,

$a(t) = \int (\sin(t))dt = -\cos(t) + C$ .

We can solve for the value of by using the acceleration given to us at t = 4 .

$a(4) = -\cos(4) + C = 2$

Therefore, $C = 2 + \cos(4)$ .

The acceleration equation is then,

$a(t) = -\cos(t)+2+\cos(4)$ .

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Example Question #442 : Spatial Calculus

The position of an object is given by the equation $x(t) = 2e^{11t}$ . What is the initial acceleration of the object?

Possible Answers:

121

242

Correct answer:

242

Explanation:

The acceleration of the object can be found by differentiating the position equation $x(t) = 2e^{11t}$ twice.

To differentiate this equation we can use the chain rule where if

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\times g'(x)$ .

Therefore the first derivative of the position equation, the velocity equation, is

$v(t) = x'(t) = 2e^{11t}(11) = 22e^{11t}$ .

Differentiating the equation a second time yields the acceleration equation.

$a(t) = 22e^{11t}(11) = 242e^{11t}$

The initial acceleration of the object is the acceleration of the object at time t = 0 .

$a(0) = 242e^{11(0)} = 242e^0 = 242(1) = 242$

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Example Question #445 : Spatial Calculus

The velocity equation of an object is $v(t) = -\frac{2}{t^3}$ . What is the acceleration equation of the object?

Possible Answers:

$a(t) = \frac{6}{t^4}$

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

$a(t) = \frac{3}{t^4}$

$a(t) = \frac{-6}{t^4}$

Correct answer:

$a(t) = \frac{6}{t^4}$

Explanation:

The acceleration equation of the object is the derivative of the velocity equation. To differentiate the velocity equation we can use the power rule where

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Rewriting the velocity we obtain

$v(t) = -\frac{2}{t^3} = -2t^{-3}$ .

Applying the power rule to the velocity equation gives us

$a(t) = v'(t) = -2(-3)t^{-3-1} = 6t^{-4}$ .

Therefore, the acceleration equation is

$a(t) = \frac{6}{t^4}$ .

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Example Question #446 : Spatial Calculus

The position of an object is given by the equation $x(t) = 4t^3 - 5t + 2\sin^2(t)$ . What is the equation of acceleration of this object?

Possible Answers:

$a(t) = 24t - 5 + 4\cos^2(t)$

$a(t) = 24t + 4\cos^2(t)-4\sin^2(t)$

$a(t) = 24 + 4\cos^2(t)-4\sin^2(t)$

$a(t) = 24t + 4\cos^2(t)+4\sin^2(t)$

Correct answer:

$a(t) = 24t + 4\cos^2(t)-4\sin^2(t)$

Explanation:

The equation of acceleration can be found by differentiating the position equation twice. To differentiate the position equation we can use the power rule and the chain rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$

and where if

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x)) \times g'(x)$

Applying these two rule to the position equation, we can find the velocity equation

$v(t) = 4(3)t^{(3-1)} - 5(1)t^{(1-1)} + 2(2)\sin^{(2-1)}(t)(\cos(t)) = 12t^2 - 5 +4\sin(t)\cos(t)$

To find the acceleration equation we must differentiate the velocity equation. To do this we must use the product rule for the third term in the velocity equation. The product rule is where if

$f(x) = j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$

Using this rule we can differentiate the velocity equation

$a(t) = v'(t) = 12(2)t^{(2-1)} - 0 + 4[\cos(t)\cos(t)+\sin(t)(-\sin(t))]$

$a(t) = 24t + 4\cos^2(t)-4\sin^2(t)$

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Example Question #52 : Acceleration

The position of a boat is defined by the equation $p(t)=6t^{3}+7t^{2}-2t+10$ . What is the boat's acceleration at t=2 ?

Possible Answers:

Correct answer:

Explanation:

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

Use the power rule to differentiate the position function twice.

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Since $p(t)=6t^{3}+7t^{2}-2t+10$ , $p'(t)=18t^{2}+14t-2$ and p''(t)=a(t)=36t+14 .

Plugging in t=2 , a(t)=36(2)+14=72+14=86 .

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Example Question #448 : Spatial Calculus

The position of a car is defined by the equation $p(t)=2t^{3}+4t^{2}+3t+14$ . What is the car's acceleration at t=4 ?

Possible Answers:

Correct answer:

Explanation:

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

Use the power rule to differentiate.

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Since $p(t)=2t^{3}+4t^{2}+3t+14$ , $p'(t)=6t^{2}+8t+3$ and p''(t)=a(t)=12t+8 .

Swapping in t=4 , a(t)=12(4)+8=48+8=56 .

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Example Question #442 : Calculus

The position of a particle is defined by the equation $p(t)=7t^{3}+2t^{2}+4t+5$ . What is the particle's acceleration at t=7 ?

Possible Answers:

299

292

290

295

298

Correct answer:

298

Explanation:

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

Use the power rule to differentiate.

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Since $p(t)=7t^{3}+2t^{2}+4t+5$ , $p'(t)=21t^{2}+4t+4$ and p''(t)=a(t)=42t+4 .

Swapping in t=7 , a(t)=42(7)+4=294+4=298 .

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Example Question #51 : How To Find Acceleration

The position of an object is given by the equation x(t) = 9t^3 - 2t . What is the acceleration of the object when t= 3 ?

Possible Answers:

162

117

Correct answer:

162

Explanation:

The acceleration equation can be found by differentiatind the position equation twice. The derivative can be found by using the power rule where if

$f(x) = x^n \rightarrow f'(x) = nx^{n-1}$

Using this to differentiate the position equation gives

$v(t) = 9(3)t^{3-1} - 2(1)t^{1-1} = 27 t^2 - 2$

Repeating this process we can find the acceleration equation.

$a(t) = 27(2)t^{(2-1)} - 2(0)t^{0-1} = 54t$

Using the acceleration equation, we can now find the acceleration at time t= 3 .

a(3) = 54(3) = 162

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #441 : Spatial Calculus

Example Question #51 : Acceleration

Example Question #443 : Spatial Calculus

Example Question #442 : Spatial Calculus

Example Question #445 : Spatial Calculus

Example Question #446 : Spatial Calculus

Example Question #52 : Acceleration

Example Question #448 : Spatial Calculus

Example Question #442 : Calculus

Example Question #51 : How To Find Acceleration

All Calculus 1 Resources

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