Question

A world record of 274 m/s for the fastest jet engine car has the following measurements. The driver makes two runs through the course, one in each direction. The car first travels from left to right and covers a distance of 604m in 2.19s. The car then reverses direction and covers the same distance in 2.22s. From these data, determine the average velocity for each run.

A. (v_textaverage,1 = 275.34 ,m/s, , v_textaverage,2 = 273.87 ,m/s)
B. (v_textaverage,1 = 273.87 ,m/s, , v_textaverage,2 = 275.34 ,m/s)
C. (v_textaverage,1 = 274.63 ,m/s, , v_textaverage,2 = 274.63 ,m/s)
D. (v_textaverage,1 = 274.00 ,m/s, , v_textaverage,2 = 274.00 ,m/s)

Answer 1

The average velocities for the jet engine car runs are 275.34 m/s for the first run and 272.07 m/s for the second run, which do not match the provided options exactly, but option A is the closest.

Step-by-step explanation:

The student has asked to determine the average velocity for each run of a jet engine car that made two runs over a set distance, one in each direction. To calculate the average velocity for each run, we use the formula: average velocity = distance / time taken. For the first run from left to right, the distance covered is 604 meters and the time taken is 2.19 seconds. For the second run in the reverse direction, the same distance is covered in 2.22 seconds. Therefore,

• Average velocity for the first run (v_textaverage,1) = distance / time = 604 m / 2.19 s = 275.34 m/s
• Average velocity for the second run (v_textaverage,2) = distance / time = 604 m / 2.22 s = 272.07 m/s

It seems there is a typo in the second velocity provided in the options, but based on our calculations, the average velocities are 275.34 m/s for the first run and 272.07 m/s for the second run. The given options do not exactly match the calculated values, but option A presents the closest values, with the second average velocity being slightly different.