If an object with uniform acceleration (or deceleration) has a speed of $4 \frac{m}{s}$ $4 m/s$ at $t = 0$ $t=0$ and moves a total of 26 m by $t = 9$ $t=9$ , what was the object's rate of acceleration?

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If an object with uniform acceleration (or deceleration) has a speed of $4 \frac{m}{s}$ $4 m/s$ at $t = 0$ $t=0$ and moves a total of 26 m by $t = 9$ $t=9$ , what was the object's rate of acceleration?

1 Answer

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Answer 1 · 2016-05-16T11:38:41

$γ = - \frac{20}{81}$ $gamma=-20/81$

Explanation:

The covered space for an uniform accelerated movement is given by
$s = s_0+v_0t+1/2 gamma t²$
In $t=0$ the movement begins with $s_0$ and $v_0=4$
After $t=9$ the object situation is
$26 = 4 * 9 + 1/2 gamma 9^2$ . Solving for $gamma=-20/81$

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If an object with uniform acceleration (or deceleration) has a speed of $4 \frac{m}{s}$ $4 m/s$ at $t = 0$ $t=0$ and moves a total of 26 m by $t = 9$ $t=9$ , 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 4 m/s4ms4 m/s at t=0t=0t=0 and moves a total of 26 m by t=9t=9t=9, what was the object's rate of acceleration?

1 Answer

Explanation:

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If an object with uniform acceleration (or deceleration) has a speed of $4 \frac{m}{s}$ $4 m/s$ at $t = 0$ $t=0$ and moves a total of 26 m by $t = 9$ $t=9$ , what was the object's rate of acceleration?