What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $12 \frac{m}{s}$ $12 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t^{2} - 3 t$ $a(t) =5t^2-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $12 \frac{m}{s}$ $12 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t^{2} - 3 t$ $a(t) =5t^2-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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Answer 1 · 2018-02-12T17:04:32

Given:
Initial velocity of the object $v 0 = 12 \frac{m}{s}$ $v0=12m/s$ at $t = 0$ $t=0$
To determine the average speed for $t (0, 5)$ $t(0,5)$
The acceleration is given by
$a (t) = 5 t^{2} - 3 t$ $a(t)=5t^2-3t$
We know that
$a (t) = \frac{d v}{d t}$ $a(t)=(dv)/dt$
Separating variables,
$d v = a (t) d t$ $dv=a(t)dt$
Substituting for $a (t)$ $a(t)$
$d v = (5 t^{2} - 3 t) d t$ $dv=(5t^2-3t)dt$
Integrating both sides
$\int d v = \int (5 t^{2} - 3 t) d t$ $intdv=int(5t^2-3t)dt$
$v = \frac{5}{3} t^{3} - \frac{3}{2} t^{2} + v 0$ $v=5/3t^3-3/2t^2+v0$

At $t = 0, v 0 = 12 \frac{m}{s}$ $t=0, v0=12m/s$
Substituting
$12 = \frac{5}{3} {(0)}^{3} - \frac{3}{2} {(0)}^{2} + v 0$ $12=5/3(0)^3-3/2(0)^2+v0$
$12 = 0 - 0 + v 0$ $12=0-0+v0$
$v 0 = 12$ $v0=12$
Thus, the expression for the velocity given by
$v (t) = \frac{5}{3} t^{3} - \frac{3}{2} t^{2} + 12$ $v(t)=5/3t^3-3/2t^2+12$
Velocity at 4 sec is
$v (4) = \frac{5}{3} {(4)}^{3} - \frac{3}{2} {(4)}^{2} + 12$ $v(4)=5/3(4)^3-3/2(4)^2+12$
$v (4) = 106.667 - 37.5 + 12$ $v(4)=106.667-37.5+12$
$v (4) = 81.167$ $v(4)=81.167$ m/s
Average velocity for the interval is
Distance travelled is obtained by integrating the expression of velocity
$v = \frac{d s}{d t}$ $v=(ds)/dt$
Separating variables,
$d s = v (t) d t$ $ds=v(t)dt$
Substituting for $v (t)$ $v(t)$
$d v = \frac{5}{3} t^{3} - \frac{3}{2} t^{2} + 12 d t$ $dv=5/3t^3-3/2t^2+12dt$

Integrating both sides
$\int d v = \int (\frac{5}{3} t^{3} - \frac{3}{2} t^{2} + 12) d t$ $intdv=int(5/3t^3-3/2t^2+12)dt$
$s (t) = \frac{5}{12} t^{4} - \frac{1}{2} t^{3} + 12 t + s 0$ $s(t)=5/12t^4-1/2t^3+12t+s0$
Assuming the object starts from rest, s0=0
$s (t) = \frac{5}{12} t^{4} - \frac{1}{2} t^{3} + 12 t + s 0$ $s(t)=5/12t^4-1/2t^3+12t+s0$
At t-4s, the position is
$s (4) = \frac{5}{12} {(4)}^{4} - \frac{1}{2} {(4)}^{3} + 12 t + 0$ $s(4)=5/12(4)^4-1/2(4)^3+12t+0$
$s (4) = 122.667$ $s(4)=122.667$

Total distance travelled in the interval of 4 seconds is $122.667$ $122.667$ m
Total time elapsed is 4 seconds

average speed = distance /time
$= \frac{122.667}{4} = 30.667 \frac{m}{s}$ $=122.667/4=30.667 m/s$

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What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $12 \frac{m}{s}$ $12 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t^{2} - 3 t$ $a(t) =5t^2-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $12 \frac{m}{s}$ $12 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t^{2} - 3 t$ $a(t) =5t^2-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?