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If a falling rock undergoes a Delta v of -98(m)/(s) due to an bar (a) of -9.8 m/s^2 how long did it fall? a 9.8 seconds b 20 seconds c 10 seconds d 5 seconds

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If a falling rock undergoes a Delta v of -98(m)/(s) due to an bar (a) of -9.8 m/s^2 how long did it fall? a 9.8 seconds b 20 seconds c 10 seconds d 5 seconds

If a falling rock undergoes a Delta v of -98(m)/(s) due to an bar (a) of -9.8 m/s^2 how long did it fall? a 9.8 seconds b 20 seconds c 10 seconds d 5 seconds

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HannahProfessional · Tutor for 6 years

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### c. 10 seconds

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## Step1: Identify the given values<br />### We are given the change in velocity ($\Delta v$) and the acceleration ($\bar{a}$). Specifically, $\Delta v = -98 \frac{m}{s}$ and $\bar{a} = -9.8 \frac{m}{s^2}$.<br />## Step2: Use the kinematic equation<br />### The kinematic equation that relates change in velocity, acceleration, and time is $\Delta v = \bar{a} \cdot t$. We need to solve for time ($t$).<br />## Step3: Solve for time<br />### Rearrange the equation to solve for $t$: <br />\[t = \frac{\Delta v}{\bar{a}}\]<br />Substitute the given values:<br />\[t = \frac{-98 \frac{m}{s}}{-9.8 \frac{m}{s^2}} = 10 \text{ seconds}\]
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