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 45 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 $4 \frac{m}{s}$ $4 m/s$ at $t = 0$ $t=0$ and moves a total of 45 m by $t = 3$ $t=3$ , what was the object's rate of acceleration?

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Answer 1 · 2015-12-30T11:23:10

$a = \frac{22}{3} \frac{m}{s^{2}}$ $a=22/3m/s^2$

Explanation:

Data:-
Initial velocity $= v_{i} = 4 \frac{m}{s}$ $=v_i=4m/s$
Distance $= S = 45 m$ $=S=45 m$
Time taken $= t = t_{2} - t_{1} = 3 - 0 = 3 s$ $=t=t_2-t_1=3-0=3s$
Acceleration $= a = ? ?$ $=a=??$
Sol:-
We know that:-
$S = v_{i} \cdot t + \frac{1}{2} a t^{2}$ $S=v_i*t+1/2at^2$
$\Rightarrow 45 = 4 \cdot 3 + \frac{1}{2} a 3^{2}$ $implies 45=4*3+1/2a3^2$
$\Rightarrow 45 = 12 + \frac{9}{2} a$ $implies 45=12+9/2a$
$\Rightarrow \frac{9}{2} a = 33 \Rightarrow 9 a = 66 \Rightarrow a = \frac{66}{9} = \frac{22}{3} \frac{m}{s^{2}}$ $implies 9/2a=33 implies 9a=66 implies a=66/9=22/3 m/s^2$
$\Rightarrow a = \frac{22}{3} \frac{m}{s^{2}}$ $implies a=22/3m/s^2$

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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 45 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 4 m/s4ms4 m/s at t=0t=0t=0 and moves a total of 45 m by t=3t=3t=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 $4 \frac{m}{s}$ $4 m/s$ at $t = 0$ $t=0$ and moves a total of 45 m by $t = 3$ $t=3$ , what was the object's rate of acceleration?