(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when
and
, and because the velocity function is continuous, you need only check the sign of
for values on the intervals (0, 3) and (3, 6).
We have, for instance
and
, which means the particle is moving the positive direction for
, or the interval (3, 6).
(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:
![\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E6%7Cv%28t%29%7C%5C%2C%5Cmathrm%20dt%3D%5Cint_0%5E3-v%28t%29%5C%2C%5Cmathrm%20dt%2B%5Cint_3%5E6v%28t%29%5C%2C%5Cmathrm%20dt)
which follows from the definition of absolute value. In particular, if
is negative, then
.
The total distance traveled is then 4 ft.
(g) Acceleration is the rate of change of velocity, so
is the derivative of
:
![a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)](https://tex.z-dn.net/?f=a%28t%29%3Dv%27%28t%29%3D-%5Cdfrac%7B%5Cpi%5E2%7D9%5Ccos%5Cleft%28%5Cdfrac%7B%5Cpi%20t%7D3%5Cright%29)
Compute the acceleration at
seconds:
![a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}](https://tex.z-dn.net/?f=a%28t%29%3D%5Cdfrac%7B%5Cpi%5E2%7D%7B18%7D%5Cdfrac%7B%5Crm%20ft%7D%7B%5Cmathrm%20s%5E2%7D)
(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)
The answer is 3x+15
You multiply the x and 5 by 3 to get 3x and 15
Answer:
Going horizontally,
Q1 a) x = 133°
Q1 b) x = 59°
Q1 c) x = 189°
Q1 d) x = 32°
Q1 e) x = 72°
Q1 f) x = 36°
Q2 a) x = 53°
Q2 b) x = 94°
Q2 c) x = 10°
Workings out:
To work out the interior angles, you need to know that angles on a straight line add up to 180°. In addition, you also need to know that angles around a point add up to 360°. When you need to find a missing angle, if the angle is on a line or in a triangle, take whatever value/values the angle/angles you have are and take it away from 180°. If the angle is around a point, (or in a square, where all angles are the same anyway) add however many values you have for the angles then take that away from 360°. Hope this helps! :)