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

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

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Answer 1 · 2017-05-16T02:37:51

$5.5 \frac{m}{s}$ $5.5 m/s$

Explanation:

I'm assuming this is related to one-dimensional motion. Combining the integral functions for velocity and position, the equation for position with respect to time is represented by

$x = x_{0} + \int_{0}^{t} [v_{0 x} + \int_{0}^{t} a_{x} d t] d t$ $x = x_0 + int_0^t [v_(0x) + int_0^ta_xdt]dt$

So, since the initial velocity $v_{0 x} = 4 \frac{m}{s}$ $v_(0x) = 4 m/s$ , the postion equation with respect to time from this is

$x = 4 \frac{m}{s} (t) + 1 \frac{m}{s^{2}} {(t)}^{2} - \frac{1}{6} \frac{m}{s^{3}} {(t)}^{3}$ $x = 4m/s(t) + 1m/(s^2)(t)^2 - 1/6m/(s^3)(t)^3$

and thus the position of the object at time $t = 3$ $t = 3$ is

$x = 4 \frac{m}{s} (3) + 1 \frac{m}{s^{2}} {(3)}^{2} - \frac{1}{6} \frac{m}{s^{3}} {(3)}^{3} = 16.5 m$ $x = 4m/s(3) + 1m/(s^2)(3)^2 - 1/6m/(s^3)(3)^3 = 16.5m$

The average velocity on the interval $t \in [0, 3]$ $t in [0,3]$ is thus

$v_(av-x) = (Deltax)/(Deltat) = (16.5m)/(3s) = 5.5 m/s$

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

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What is the average speed of an object that is moving at 4 m/s4ms4 m/s at t=0t=0t=0 and accelerates at a rate of a(t) =2-ta(t)=2−ta(t) =2-t on t in [0,3]t∈[0,3]t in [0,3]?

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