If an object with uniform acceleration (or deceleration) has a speed of $3 \frac{m}{s}$ $3 m/s$ at $t = 0$ $t=0$ and moves a total of 58 m by $t = 6$ $t=6$ , what was the object's rate of acceleration?

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If an object with uniform acceleration (or deceleration) has a speed of $3 \frac{m}{s}$ $3 m/s$ at $t = 0$ $t=0$ and moves a total of 58 m by $t = 6$ $t=6$ , what was the object's rate of acceleration?

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Answer 1 · 2016-02-21T14:39:06

a=20/9 m/s^2
a=-20/9 m/s^2

Explanation:

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$area under graph is equal displacement of 58 m$ $"area under graph is equal displacement of 58 m"$
$58 = \frac{3 + k}{2} \cdot 6; 58 = (3 + k) \cdot 3; 58 = 9 + 3 \cdot k$ $58=(3+k)/2*6"; "58=(3+k)*3"; "58=9+3*k$
$49 = 3 \cdot k; k = \frac{49}{3}$ $49=3*k"; "k=49/3$
$rate of acceleration is equal tangent of graph$ $"rate of acceleration is equal tangent of graph"$
$a = \frac{3 - k}{6} or a = \frac{k - 3}{a}$ $a=(3-k)/6" or " a=(k-3)/a$
$a = \frac{3 - \frac{49}{3}}{6}; a = - \frac{20}{9} \frac{m}{s^{2}}$ $a=(3-49/3)/6" ; " a=-20/9 m/s^2$
$a = \frac{\frac{49}{3} - 3}{6}; a = \frac{20}{9} \frac{m}{s^{2}}$ $a=(49/3-3)/6" ; " a=20/9 m/s^2$

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If an object with uniform acceleration (or deceleration) has a speed of $3 \frac{m}{s}$ $3 m/s$ at $t = 0$ $t=0$ and moves a total of 58 m by $t = 6$ $t=6$ , what was the object's rate of acceleration?

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If an object with uniform acceleration (or deceleration) has a speed of 3ms3 m/s at t=0t=0 and moves a total of 58 m by t=6t=6, what was the object's rate of acceleration?

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If an object with uniform acceleration (or deceleration) has a speed of $3 \frac{m}{s}$ $3 m/s$ at $t = 0$ $t=0$ and moves a total of 58 m by $t = 6$ $t=6$ , what was the object's rate of acceleration?