What is the average speed 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 $a (t) = 5 - 3 t$ $a(t) =5-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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Answer 1 · 2017-05-04T10:10:06

The average speed is $= 6.5 m s^{- 1}$ $=6.5ms^-1$

The speed is the integral of the acceleration

$a (t) = 5 - 3 t$ $a(t)=5-3t$

$v (t) = \int (5 - 3 t) d t$ $v(t)=int(5-3t)dt$

$v (t) = 5 t - \frac{3}{2} t^{2} + C$ $v(t)=5t-3/2t^2+C$

Plugging in the initial conditions

$v (0) = 0 - 0 + C = 12$ $v(0)=0-0+C=12$

So,

$C = 12$ $C=12$

and

$v (t) = 5 t - \frac{3}{2} t^{2} + 12$ $v(t)=5t-3/2t^2+12$

The average speed is

$4 ¯ v = \int_{0}^{4} (5 t - \frac{3}{2} t^{2} + 12) d t$ $4barv=int_0^4(5t-3/2t^2+12)dt$

$= {[\frac{5}{2} t^{2} - \frac{3}{6} t^{3} + 12 t]}_{0}^{2}$ $=[5/2t^2-3/6t^3+12t]_0^4$

$= (10 - 32 + 48) - (0)$ $=(10-32+48)-(0)$

$= 26$ $=26$

$¯ v = \frac{26}{4} = 6.5 m s^{- 1}$ $barv=26/4=6.5ms^-1$

What is the average speed of an object that is moving at 12ms12 m/s at t=0t=0 and accelerates at a rate of a(t)=5−3ta(t) =5-3t on t∈[0,4]t in [0,4]?