Question

A motorcyclist who is moving along an x axis directed toward the east has an acceleration given by a = (6.30 - 2.20t) m/s2 for 0 ≤ t ≤ 6.0 s. At t = 0, the velocity and position of the cyclist are 3.00 m/s and 7.3 m. What total distance does the cyclist travel between t = 0 and 6.0 s? idk why but i keep getting the wrong answer.

Answer 1

What answer do you come up with?

Answer 2

51.5

Answer 3

we have to integrate the acceleration first, in order to get the velosity function:

\[v\left( t \right) = \int {a\left( t \right)dt} = \int {\left( {6.3 - 2.2t} \right)dt} = ...?\]

Answer 4

oops... the velocity function...

Answer 5

so it would be 6.3t - (2.2/2)t^2 ?

Answer 6

we have to add the arbitrary constant, namely:
\[v\left( t \right) = 6.3t - 1.1{t^2} + C\]

Answer 7

next, we have to apply the initial condition, nameli v(0) = 3. So we get:
\[\begin{gathered}
3 = v\left( 0 \right) = 6.3 \cdot 0 - 1.1 \cdot 0 + C = C \hfill \\
C = 3 \hfill \\
\end{gathered} \]

Answer 8

finally our velocity function will be this:
\[v\left( t \right) = 6.3t - 1.1{t^2} + 3\]

Answer 9

and then put t = 6 in the equation?

Answer 10

so the answer is 1.2?

Answer 11

no, we have to integrate the velocity function, in order to get the position function x(t), namely:
\[x\left( t \right) = \int {\left( {6.3t - 1.1{t^2} + 3} \right)dt} \]

Answer 12

oh the answer is 37.2?

Answer 13

you should get this:
\[x\left( t \right) = \int {\left( {6.3t - 1.1{t^2} + 3} \right)dt} = \frac{{6.3}}{2}{t^2} - \frac{{1.1}}{3}{t^3} + 3t + {C_1}\]

Answer 14

where C_1 is another arbitrary constant, whose value is given applying the initial condition for the position x, namely:
\[\begin{gathered}
7.3 = x\left( 0 \right) = \frac{{6.3}}{2} \cdot 0 - \frac{{1.1}}{3} \cdot 0 + 3 \cdot 0 + {C_1} = {C_1} \hfill \\
{C_1} = 7.3 \hfill \\
\end{gathered} \]

Answer 15

finally, the position function, will be this:
\[x\left( t \right) = \frac{{6.3}}{2}{t^2} - \frac{{1.1}}{3}{t^3} + 3t + 7.3\]

Answer 16

The requested distance is therefore:
\[\begin{gathered}
x\left( 6 \right) - x\left( 0 \right) = \hfill \\
= \left[ {\frac{{6.3}}{2} \cdot {6^2} - \frac{{1.1}}{3} \cdot {6^3} + 3 \cdot 6 + 7.3} \right] - \left[ {\frac{{6.3}}{2} \cdot 0 - \frac{{1.1}}{3} \cdot 0 + 3 \cdot 0 + 7.3} \right] = ...? \hfill \\
\end{gathered} \]

Answer 17

-675?

Answer 18

I got 52.2, is it right?

Answer 19

yes! thank you :)

Answer 20

thank you!