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xxMikexx [17]
3 years ago
11

A particle sits on a smooth surface and is acted upon by a time dependent horizontal force, giving it an

Mathematics
1 answer:
garik1379 [7]3 years ago
7 0

(a) By the fundamental theorem of calculus,

<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>

The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives

<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²

(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of

∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024

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Answer:

\frac{\pi  }{3}

2 * \pi * 3  * \frac{20}{360}

120\pi/360 = \frac{\pi  }{3}

Step-by-step explanation:

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Answer:

m∠ABC = 74

Step-by-step explanation:

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On the 1st January 2014 Carol invested some money in a bank account.
Ghella [55]

Answer:

\large \boxed{\text{\pounds 23 360.00}}

Step-by-step explanation:

The formula for the accrued amount from compound interest is

A = P \left(1 + \dfrac{r}{n}\right)^{nt}

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:  

r = 0.025

\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}

3. Summary

1 Jan 2014      P = £23 360.00

1 Jan 2015     A =    23 944.00

     Withdrawal = <u>    -1  000.00 </u>

                     P =     22 944.00

1 Jan 2016    A =    £23 517.60

5 0
3 years ago
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