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# 65165

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A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far has the boat traveled? Round your answer to the nearest whole number.

Jan 14, 2016

#5
+15

Hi guest and Solveit,

A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far has the boat traveled? Round your answer to the nearest whole number.

Your formula is incorrect guest .    The one you are thinking of is   v=u+at

u=2.5m/s,     a=4.2m/s^2      t=6s,      find s

The one you need is

$$s=ut+0.5at^2\\ s=2.5*6+\frac{1}{2}*4.2*36\\ s=15+2.1*36\\ s=90.6\; metres.$$

When I was at school I did both maths and physics.

Now in physics I used the physics formulas    BUT     in maths ir was compulsory to use calculus!

So now I will do it with calculus.

$$x\; is \;displacement,\;\;\\\frac{dx}{dt}\;\;is \;velocity\;\;\;\\\frac{d^2x}{dt^2}\;\;is \;acceleration\;\;\;and\;\\t\;is\;time.\\$$

$$\frac{d^2x}{dt^2}=4.2\\ \frac{dx}{dt}=4.2t+c_1\\ when\;t=0\;\;\frac{dx}{dt}=2.5\\ \quad sub\; in\\ 2.5=4.2*0+c_1\\ c_1=2.5\\ \frac{dx}{dt}=4.2t+2.5\\ x=\frac{4.2t^2}{2}+2.5t+c_2\\ \mbox{When t=0, x=0 so }c_2=0\\ x=\frac{4.2t^2}{2}+2.5t\\ When \; t=6\\ x=2.1*6^2+2.5*6\\ x=90.6\;metres$$

That is how you do it using calculus Jan 15, 2016

#1
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i think:

4.2-2.5 = 1.7      2.5+0.85 = 3.35 it is his avareage speed

3.35*6 = 20.1

Jan 14, 2016
#2
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Simple SUVAT problem,

S = u + at
S = 2.5 + 4.2*6

S = 27.2 meters

Jan 14, 2016
#3
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doesn t it saying that initial speed was 2.5 and it speeded up to 4.2 in 6s

Jan 14, 2016
#4
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By the units given, the initial speed is 2.5 ms^-1 where it's acceleration is 4.2 ms^-2.

Jan 14, 2016
#5
+15

Hi guest and Solveit,

A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far has the boat traveled? Round your answer to the nearest whole number.

Your formula is incorrect guest .    The one you are thinking of is   v=u+at

u=2.5m/s,     a=4.2m/s^2      t=6s,      find s

The one you need is

$$s=ut+0.5at^2\\ s=2.5*6+\frac{1}{2}*4.2*36\\ s=15+2.1*36\\ s=90.6\; metres.$$

When I was at school I did both maths and physics.

Now in physics I used the physics formulas    BUT     in maths ir was compulsory to use calculus!

So now I will do it with calculus.

$$x\; is \;displacement,\;\;\\\frac{dx}{dt}\;\;is \;velocity\;\;\;\\\frac{d^2x}{dt^2}\;\;is \;acceleration\;\;\;and\;\\t\;is\;time.\\$$

$$\frac{d^2x}{dt^2}=4.2\\ \frac{dx}{dt}=4.2t+c_1\\ when\;t=0\;\;\frac{dx}{dt}=2.5\\ \quad sub\; in\\ 2.5=4.2*0+c_1\\ c_1=2.5\\ \frac{dx}{dt}=4.2t+2.5\\ x=\frac{4.2t^2}{2}+2.5t+c_2\\ \mbox{When t=0, x=0 so }c_2=0\\ x=\frac{4.2t^2}{2}+2.5t\\ When \; t=6\\ x=2.1*6^2+2.5*6\\ x=90.6\;metres$$

That is how you do it using calculus Melody Jan 15, 2016
#6
+5

Nice, Melody......it's interesting to see how the Physics "formulas" can be derived using Calculus.......   Jan 15, 2016
#7
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Thanks Chris :D

Jan 15, 2016