An object is moving at a speed of 4 inches every 5 minutes. Express this speed in centimeters per second. Round your answer to the nearest hundredth.
*Note: you must use these exact conversion factors to get this question right.
\begin{tabular}{|l|l|}
\hline Distance / length & Time \\
\hline 1 foot (ft) = 12 inches (in) & 1 minute $( min )=60$ seconds $( sec )$ \\
\hline 1 yard $( yd )=3$ feet $( ft )$ & 1 hour ( hr ) $=60$ minutes (min) \\
\hline 1 mile $( mi )=5280$ feet $( ft )$ & 1 day (day) $=24$ hours (hr) \\
\hline 1 meter (m) = 100 centimeters (cm) & 1 week (week) = 7 days (days) \\
\hline 1 kilometer (km) = 1000 meters (m) & 1 month (month) = 30 days (days) \\
\hline 1 inch (in) $=2.54$ centimeters (cm) & 1 year (year) = 365 days (days) \\
\hline 1 foot $( ft )=0.305$ meters (m) & \\
\hline 1 mile (mi) = 1.609 kilometers (km) & \\
\hline
\end{tabular}
Answer (3)
Well, the car gets to 5.8 3.3=19.14 m/s, getting to 5.8 3.3*3.3=63.162 meters from the start. For 9.8 secons it goes 187.572 meters, a total of 250.734. So, in 5.056 meters, it stopped uniformly. The speed dropped from 19.14 to 0 in 5.056 meters. There was a formula which didn't use time, but I've forgotten it. It would have given us the deceleration and then one could find the time through a 2nd degree equation. The timing for the first two parts is already known. Either way, we've calculate more than we need to know for this question. You can cut the answer from that 250.734 distance.
raheh ;
The blue car travels approximately 219.07 meters before applying the brakes. This distance includes the distance covered during both the acceleration and constant speed phases. The calculations involve using kinematic equations for uniform acceleration and constant speed.
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