If an object with uniform acceleration (or deceleration) has a speed of $2 \frac{m}{s}$ $2 m/s$ at $t = 0$ $t=0$ and moves a total of 90 m by $t = 6$ $t=6$ , what was the object's rate of acceleration?

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Answer 1 · 2017-05-30T06:55:35

The rate of acceleration is $= 4.33 m s^{- 2}$ $=4.33ms^-2$

The initial speed is $u = 2 m s^{- 1}$ $u=2ms^-1$

The distance is $s = 90 m$ $s=90m$

The time is $t = 6 s$ $t=6s$

The acceleration is $a = ?$ $a=?$

We apply the equation of motion

$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$

Therefore,

$\frac{1}{2} a t^{2} = s - u t$ $1/2at^2=s-ut$

$a = \frac{s - u t}{\frac{1}{2} t^{2}}$ $a=(s-ut)/(1/2t^2)$

$= \frac{90 - 6 \cdot 2}{\frac{1}{2} \cdot 6^{2}}$ $=(90-6*2)/(1/2*6^2)$

$= \frac{78}{18}$ $=78/18$

$= 4.33 m s^{- 2}$ $=4.33ms^-2$

If an object with uniform acceleration (or deceleration) has a speed of 2 m/s2ms2 m/s at t=0t=0t=0 and moves a total of 90 m by t=6t=6t=6, what was the object's rate of acceleration?