Question

An object moving with uniform acceleration has a velocity of 11.0 cm/s in the positive x-direction when its x-coordinate is 2.79 cm. If its x-coordinate 2.05 s later is −5.00 cm, what is its acceleration? ______ cm/s2

Answer 1

-14.43 cm/s^2

Explanation

to solve this we need to use the formula

$\begin{gathered} x=v_ot+\frac{1}{2}at^2 \\ where \\ v_o\text{ is the inital velocity} \\ t\text{ is the time} \\ x\text{ is the traveled distance} \\ a\text{ is the acceleration} \end{gathered}$

Step 1

given

$\begin{gathered} v_1=11\frac{cm}{s} \\ x_1=2.79\text{ cm} \\ x_2=-5\text{ cm} \\ time=2.05\text{ s} \end{gathered}$

a) find the traveled distance

$\begin{gathered} x=\Delta x=-5-(2.79)=-7.79 \\ the\text{ negative sign indicates the opposite side} \end{gathered}$

b) replace in the formula and solve for a

$\begin{gathered} x=v_{o}t+\frac{1}{2}at^{2} \\ -7.79=11\frac{cm}{s}*2.05s+\frac{1}{2}a*(2.05\text{ s\rparen}^2 \\ -7.79=22.55+2.10125a \\ subtract\text{ 22.55 in both sides} \\ -7.79-22.55=22.55+2.10125a-22.55 \\ -30.34=2.10125a \\ divide\text{ both sides by 2.10125} \\ \frac{-30.34}{2.10125}=\frac{2.101,25a}{2.10125} \\ -14.43\frac{cm}{s^2}=a \end{gathered}$

so, the acceleation is

-14.43 cm/s^2

I hope this helps you