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

Step-by-step explanation

to solve this we need to use the formula

`\begin{gathered} x=v_ot+(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(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+(1)/(2)at^(2) \\ -7.79=11(cm)/(s)*2.05s+(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} \\ (-30.34)/(2.10125)=(2.101,25a)/(2.10125) \\ -14.43(cm)/(s^2)=a \end{gathered}`

so, the acceleation is

-14.43 cm/s^2

I hope this helps you