What is the average speed of an object that is still at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t^{2} - 4$ $a(t) =t^2-4$ from $t \in [2, 3]$ $t in [2, 3]$ ?

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What is the average speed of an object that is still at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t^{2} - 4$ $a(t) =t^2-4$ from $t \in [2, 3]$ $t in [2, 3]$ ?

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Answer 1 · 2018-01-12T09:28:37

4.58 $\frac{m}{sec}$ $m/sec$

Explanation:

Acceleration of the object is given as, $a = 4 t^{2} - 4$ $a=4t^2-4$
So,velocity will be $\frac{t^{3}}{3} - 4 t$ $(t^3)/3-4t$ (by integrating)
Hence,displacement will be,
$S = \frac{t^{4}}{12} - (2 t^{2})$ $S=(t^4)/12-(2t^2)$ (by integrating)
Hence,displacement with in t=2 to t=3 will be -4.58 meter
Now here distance=displacement that has happened in between 2 sec and 3 sec
Hence average speed with in this duration=(total distance covered/total time)
I.e $(\frac{4.58}{1})$ $(4.58/1)$ $\frac{m}{sec}$ $m/sec$ or 4.58 $\frac{m}{sec}$ $m/sec$

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What is the average speed of an object that is still at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t^{2} - 4$ $a(t) =t^2-4$ from $t \in [2, 3]$ $t in [2, 3]$ ?

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What is the average speed of an object that is still at t=0t=0t=0 and accelerates at a rate of a(t) =t^2-4a(t)=t2−4a(t) =t^2-4 from t in [2, 3]t∈[2,3]t in [2, 3]?

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What is the average speed of an object that is still at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t^{2} - 4$ $a(t) =t^2-4$ from $t \in [2, 3]$ $t in [2, 3]$ ?