What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $8 \frac{m}{s}$ $8 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t^{2} - 5$ $a(t) =t^2-5$ on $t \in [0, 3]$ $t in [0,3]$ ?

Question

1488 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-25T14:53:08

The average speed is $= 3.92 m s^{- 1}$ $=3.92ms^-1$

The speed is the integral of the acceleration

$a (t) = t^{2} - 5$ $a(t)=t^2-5$

$v (t) = \frac{1}{3} t^{3} - 5 t + C$ $v(t)=1/3t^3-5t+C$

Plugging in the initial conditions

$v (0) = 8 m s^{- 1}$ $v(0)=8ms^-1$

$v (0) = 0 - 0 + C = 8$ $v(0)=0-0+C=8$

Therefore,

$v (t) = \frac{1}{3} t^{3} - 5 t + 8$ $v(t)=1/3t^3-5t+8$

$v (3) = 9 - 15 + 8 = 2$ $v(3)=9-15+8=2$

$v (4) = \frac{64}{3} - 20 + 8 = \frac{28}{3}$ $v(4)=64/3-20+8=28/3$

The average speed is

$(5 - 0) ¯ v = \int_{0}^{3} (\frac{1}{3} t^{3} - 5 t + 8) d t + v (3) + v (4)$ $(5-0)barv=int_0^3(1/3t^3-5t+8)dt+v(3)+v(4)$

$5 ¯ v = {[\frac{1}{12} t^{4} - \frac{5}{2} t^{2} + 8 t]}_{0}^{4} + 2 + \frac{28}{3}$ $5barv=[1/12t^4-5/2t^2+8t]_0^3+2+28/3$

$= \frac{27}{4} - \frac{45}{2} + 24 + \frac{34}{3}$ $=27/4-45/2+24+34/3$

$= 19.58$ $=19.58$

$¯ v = \frac{19.58}{5} = 3.92 m s^{- 1}$ $barv=19.58/5=3.92ms^-1$

What is the average speed, on t∈[0,5]t in [0,5], of an object that is moving at 8ms8 m/s at t=0t=0 and accelerates at a rate of a(t)=t2−5a(t) =t^2-5 on t∈[0,3]t in [0,3]?