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.
+118608 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 
+2498 i think:
4.2-2.5 = 1.7 2.5+0.85 = 3.35 it is his avareage speed
3.35*6 = 20.1
the answer: 20
+2498 doesn t it saying that initial speed was 2.5 and it speeded up to 4.2 in 6s
By the units given, the initial speed is 2.5 ms^-1 where it's acceleration is 4.2 ms^-2.
+118608 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
