# Acceleration question

A particle moves along the $x$-axis so that its velocity is given by $v(t)=5+3.1\cos(0.8t)$ for $t\geq 0$. What is the acceleration of the particle at $t=4$.

I differentiated this and got:

$a(t)=-2.48 \sin(0.8t)$

and plugged in $4$ and got

$a(4) = -0.138437$

But none of the answer choices fit the answer, am I wrong or is the question wrong?

Answer choices: $-0.045$, $-0.145$, $-0.181$, $0.145$, $5.145$

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$t$ is expressed in radians!! –  mfl Jun 27 at 2:34
Yes, check your calculations, making sure that your calculator is in radian mode. –  Gahawar Jun 27 at 2:35
It is... I dont know what im doing wrong –  Panthy Jun 27 at 2:35
Check again. I have obtained you result in DEG mode. –  mfl Jun 27 at 2:38
Ok my calculator is weird, i entered it on wolfram and got around 0.145 :) –  Panthy Jun 27 at 2:39

Instructions:

1. Click the start button.
2. Open calculator.
3. Click "view" and select Scientific.
6. Get $0.1447678757...$
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wolframalpha.com/input/?i=-2.48sin%280.8*4%29+in+radians why does it say its +0.145 here... which one to go with?D: –  Panthy Jun 27 at 2:40
@Panthy It is 0.145. That was a typo. SORRY! –  BlackAdder Jun 27 at 2:44
ahh thanks!.......... –  Panthy Jun 27 at 2:47

v(t)=5+3.1cos (0.8t)

Differentiating with respect to t

We get

a(t)=3.1*(-sin (0.8t))*0.8 =-2.48sin (0.8t)

a(4)=-2.48sin (3.2) =-2.48sin (3.14-3.2) =+2.48sin(0.06) =+2.48*0.06=+0.1488

Note that the answer here is a round about value as pi is approximately equal to 3.14 and sinx is approximately equal to x for x<0.5

This is the answer if you don't want to use a calculator. :)

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