What is the average speed of an object that is moving at $9 \frac{m}{s}$ $9 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 t - 5$ $a(t) =2t-5$ on $t \in [1, 4]$ $t in [1,4]$ ?

Question

1398 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-01T05:40:25

The average speed is $= 3.5 m s^{- 2}$ $=3.5ms^-2$

$a (t) = 2 t - 5$ $a(t)=2t-5$

The speed is the integral of the acceleration

$v (t) = \int (2 t - 5) d t$ $v(t)=int(2t-5)dt$

$v (t) = \frac{2}{2} t^{2} - 5 t + C$ $v(t)=2/2t^2-5t+C$

Initial conditions are,

$v (0) = 9$ $v(0)=9$

Therefore,

$v (0) = 0 - 0 + C = 9$ $v(0)=0-0+C=9$

So,

$v (t) = t^{2} - 5 t + 9$ $v(t)=t^2-5t+9$

The average velocity is $¯ v$ $barv$

$(4 - 1) ¯ v = \int_{1}^{4} (t^{2} - 5 t + 9) d t$ $(4-1)barv=int_1^4(t^2-5t+9)dt$

$3 ¯ v = {[\frac{t^{3}}{3} - \frac{5}{2} t^{2} + 9 t]}_{1}^{2}$ $3barv=[t^3/3-5/2t^2+9t]_1^4$

$= (\frac{64}{3} - 40 + 36) - (\frac{1}{3} - \frac{5}{2} + 9)$ $=(64/3-40+36)-(1/3-5/2+9)$

$3 ¯ v = \frac{52}{3} - \frac{41}{6} = \frac{63}{6} = \frac{21}{2}$ $3barv=52/3-41/6=63/6=21/2$

$¯ v = \frac{7}{2} m s^{- 1}$ $barv=7/2ms^-1$

What is the average speed of an object that is moving at 9ms9 m/s at t=0t=0 and accelerates at a rate of a(t)=2t−5a(t) =2t-5 on t∈[1,4]t in [1,4]?