QuestionFind the car's acceleration at $t=2$ seconds given $s(t)=2 t^{4}-10 t^{3}-5 t^{2}+9$ . Answer in $\mathrm{m} / \mathrm{s}^{2}$ .

Studdy Solution

STEP 1

Assumptions1. The position of the car is given by the function $s(t)= t^{4}-10 t^{3}-5 t^{}+9$ . We need to find the acceleration of the car at $t=$ seconds3. The acceleration is the second derivative of the position function

STEP 2

First, we need to find the velocity of the car, which is the first derivative of the position function.
$v(t) = \frac{ds}{dt}$

STEP 3

Now, calculate the first derivative of the position function.
$v(t) = \frac{d}{dt}(2 t^{}-10 t^{3}-5 t^{2}+9)$

STEP 4

Apply the power rule for differentiation to each term of the function.
$v(t) =8t^{3} -30t^{2} -10t$

STEP 5

Now, we need to find the acceleration of the car, which is the second derivative of the position function, or the first derivative of the velocity function.
$a(t) = \frac{dv}{dt}$

STEP 6

Now, calculate the first derivative of the velocity function.
$a(t) = \frac{d}{dt}(8t^{3} -30t^{2} -10t)$

STEP 7

Apply the power rule for differentiation to each term of the function.
$a(t) =24t^{2} -60t -10$

STEP 8

Now that we have the acceleration function, we can find the acceleration of the car at $t=2$ seconds by plugging $t=2$ into the acceleration function.
$a(2) =24(2)^{2} -60(2) -10$

STEP 9

Calculate the acceleration at $t=2$ seconds.
$a(2) =24(4) -60(2) - =96 -120 - = -34 \, \mathrm{m/s}^{2}$ The acceleration of the car at $t=2$ seconds is $-34 \, \mathrm{m/s}^{2}$ .

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QuestionFind the car's acceleration at $t=2$ seconds given $s(t)=2 t^{4}-10 t^{3}-5 t^{2}+9$ . Answer in $\mathrm{m} / \mathrm{s}^{2}$ .

Studdy Solution

STEP 1

Assumptions1. The position of the car is given by the function $s(t)= t^{4}-10 t^{3}-5 t^{}+9$ . We need to find the acceleration of the car at $t=$ seconds3. The acceleration is the second derivative of the position function

STEP 2

First, we need to find the velocity of the car, which is the first derivative of the position function.
$v(t) = \frac{ds}{dt}$

STEP 3

Now, calculate the first derivative of the position function.
$v(t) = \frac{d}{dt}(2 t^{}-10 t^{3}-5 t^{2}+9)$

STEP 4

Apply the power rule for differentiation to each term of the function.
$v(t) =8t^{3} -30t^{2} -10t$

STEP 5

Now, we need to find the acceleration of the car, which is the second derivative of the position function, or the first derivative of the velocity function.
$a(t) = \frac{dv}{dt}$

STEP 6

Now, calculate the first derivative of the velocity function.
$a(t) = \frac{d}{dt}(8t^{3} -30t^{2} -10t)$

STEP 7

Apply the power rule for differentiation to each term of the function.
$a(t) =24t^{2} -60t -10$

STEP 8

Now that we have the acceleration function, we can find the acceleration of the car at $t=2$ seconds by plugging $t=2$ into the acceleration function.
$a(2) =24(2)^{2} -60(2) -10$

STEP 9

Calculate the acceleration at $t=2$ seconds.
$a(2) =24(4) -60(2) - =96 -120 - = -34 \, \mathrm{m/s}^{2}$ The acceleration of the car at $t=2$ seconds is $-34 \, \mathrm{m/s}^{2}$ .

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QuestionFind the car's acceleration at t=2t=2t=2 seconds given s(t)=2t4−10t3−5t2+9s(t)=2 t^{4}-10 t^{3}-5 t^{2}+9s(t)=2t4−10t3−5t2+9. Answer in m/s2\mathrm{m} / \mathrm{s}^{2}m/s2.

Studdy Solution

STEP 1

Assumptions1. The position of the car is given by the function s(t)=t4−10t3−5t+9s(t)= t^{4}-10 t^{3}-5 t^{}+9s(t)=t4−10t3−5t+9 . We need to find the acceleration of the car at t=t=t= seconds3. The acceleration is the second derivative of the position function

STEP 2

First, we need to find the velocity of the car, which is the first derivative of the position function.v(t)=dsdtv(t) = \frac{ds}{dt}v(t)=dtds​

STEP 3

Now, calculate the first derivative of the position function.v(t)=ddt(2t−10t3−5t2+9)v(t) = \frac{d}{dt}(2 t^{}-10 t^{3}-5 t^{2}+9)v(t)=dtd​(2t−10t3−5t2+9)

STEP 4

Apply the power rule for differentiation to each term of the function.v(t)=8t3−30t2−10tv(t) =8t^{3} -30t^{2} -10tv(t)=8t3−30t2−10t

STEP 5

Now, we need to find the acceleration of the car, which is the second derivative of the position function, or the first derivative of the velocity function.a(t)=dvdta(t) = \frac{dv}{dt}a(t)=dtdv​

STEP 6

Now, calculate the first derivative of the velocity function.a(t)=ddt(8t3−30t2−10t)a(t) = \frac{d}{dt}(8t^{3} -30t^{2} -10t)a(t)=dtd​(8t3−30t2−10t)

STEP 7

Apply the power rule for differentiation to each term of the function.a(t)=24t2−60t−10a(t) =24t^{2} -60t -10a(t)=24t2−60t−10

STEP 8

Now that we have the acceleration function, we can find the acceleration of the car at t=2t=2t=2 seconds by plugging t=2t=2t=2 into the acceleration function.a(2)=24(2)2−60(2)−10a(2) =24(2)^{2} -60(2) -10a(2)=24(2)2−60(2)−10

STEP 9

Calculate the acceleration at t=2t=2t=2 seconds.a(2)=24(4)−60(2)−=96−120−=−34 m/s2a(2) =24(4) -60(2) - =96 -120 - = -34 \, \mathrm{m/s}^{2}a(2)=24(4)−60(2)−=96−120−=−34m/s2The acceleration of the car at t=2t=2t=2 seconds is −34 m/s2-34 \, \mathrm{m/s}^{2}−34m/s2.

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QuestionFind the car's acceleration at $t=2$ seconds given $s(t)=2 t^{4}-10 t^{3}-5 t^{2}+9$ . Answer in $\mathrm{m} / \mathrm{s}^{2}$ .

Assumptions1. The position of the car is given by the function $s(t)= t^{4}-10 t^{3}-5 t^{}+9$ . We need to find the acceleration of the car at $t=$ seconds3. The acceleration is the second derivative of the position function

First, we need to find the velocity of the car, which is the first derivative of the position function.
$v(t) = \frac{ds}{dt}$

Now, calculate the first derivative of the position function.
$v(t) = \frac{d}{dt}(2 t^{}-10 t^{3}-5 t^{2}+9)$

Apply the power rule for differentiation to each term of the function.
$v(t) =8t^{3} -30t^{2} -10t$

Now, we need to find the acceleration of the car, which is the second derivative of the position function, or the first derivative of the velocity function.
$a(t) = \frac{dv}{dt}$

Now, calculate the first derivative of the velocity function.
$a(t) = \frac{d}{dt}(8t^{3} -30t^{2} -10t)$

Apply the power rule for differentiation to each term of the function.
$a(t) =24t^{2} -60t -10$

Now that we have the acceleration function, we can find the acceleration of the car at $t=2$ seconds by plugging $t=2$ into the acceleration function.
$a(2) =24(2)^{2} -60(2) -10$

Calculate the acceleration at $t=2$ seconds.
$a(2) =24(4) -60(2) - =96 -120 - = -34 \, \mathrm{m/s}^{2}$ The acceleration of the car at $t=2$ seconds is $-34 \, \mathrm{m/s}^{2}$ .