If an object with uniform acceleration (or deceleration) has a speed of $9 \frac{m}{s}$ $9 m/s$ at $t = 0$ $t=0$ and moves a total of 44 m by $t = 3$ $t=3$ , what was the object's rate of acceleration?

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If an object with uniform acceleration (or deceleration) has a speed of $9 \frac{m}{s}$ $9 m/s$ at $t = 0$ $t=0$ and moves a total of 44 m by $t = 3$ $t=3$ , what was the object's rate of acceleration?

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Answer 1 · 2016-09-01T05:06:38

$3 . . 7 m s^{- 2}$ $3.dot7ms^-2$

Explanation:

Given distance moved $s = 44 m$ $s=44m$ in time $3 s$ $3s$ , initial speed at $t = 0$ $t=0$ is $9 m s^{- 1}$ $9ms^-1$ . To find acceleration $a$ $a$ .
We can use kinematic equation connecting above variables
$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$
Plugin given values
$44 = 9 \times 3 + \frac{1}{2} a {(3)}^{2}$ $44=9xx3+1/2a(3)^2$
$\Rightarrow \frac{9}{2} a = 44 - 27$ $=>9/2a=44-27$
$\Rightarrow \frac{9}{2} a = 44 - 27$ $=>9/2a=44-27$
$\Rightarrow a = 3 . . 7 m s^{- 2}$ $=>a=3.dot7ms^-2$

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If an object with uniform acceleration (or deceleration) has a speed of $9 \frac{m}{s}$ $9 m/s$ at $t = 0$ $t=0$ and moves a total of 44 m by $t = 3$ $t=3$ , what was the object's rate of acceleration?

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If an object with uniform acceleration (or deceleration) has a speed of 9 m/s9ms9 m/s at t=0t=0t=0 and moves a total of 44 m by t=3t=3t=3, what was the object's rate of acceleration?

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Explanation:

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If an object with uniform acceleration (or deceleration) has a speed of $9 \frac{m}{s}$ $9 m/s$ at $t = 0$ $t=0$ and moves a total of 44 m by $t = 3$ $t=3$ , what was the object's rate of acceleration?