1. What is the shortest possible time in which a bacterium could travel a distance of 8.4 cm across a Petri dish at a constant speed of 3.5mm/s 2. A child is pushing a shopping cart at a speed of 1.5m/s How long will it take this child to push the cart down an aisle with a length of 9.3 m? 3. An athlete swims from the north end to the south end of a 50.0 m pool in 20.0 s and makes the return trip to the starting position in 22.0 s. a. What is the average velocity for the first half of the swim? b. What is the average velocity for the second half of the swim? c. What is the average velocity for the roundtrip? 4. Two students walk in the same direction along a straight path at a con- stant speed-one at 0.90m/s and the other at 1.90m/s a. Assuming that they start at the same point and the same time,how much sooner does the faster student arrive at a destination 780 m away? b. How far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

1. To find the shortest possible time in which a bacterium could travel a distance of 8.4 cm across a Petri dish at a constant speed of 3.5 mm/s, we can use the formula:<br /><br />Time = Distance / Speed<br /><br />First, we need to convert the distance to the same unit as the speed. Since the speed is given in mm/s, we'll convert the distance to mm:<br /><br />8.4 cm = 84 mm<br /><br />Now we can calculate the time:<br /><br />Time = 84 mm / 3.5 mm/s = 24 seconds<br /><br />Therefore, the shortest possible time in which the bacterium could travel a distance of 8.4 cm across the Petri dish at a constant speed of 3.5 mm/s is 24 seconds.<br /><br />2. To find how long it will take the child to push the cart down an aisle with a length of 9.3 m at a speed of 1.5 m/s, we can use the formula:<br /><br />Time = Distance / Speed<br /><br />Using the given values, we can calculate the time:<br /><br />Time = 9.3 m / 1.5 m/s = 6.2 seconds<br /><br />Therefore, it will take the child 6.2 seconds to push the cart down an aisle with a length of 9.3 m at a speed of 1.5 m/s.<br /><br />3. a. To find the average velocity for the first half of the swim, we can use the formula:<br /><br />Average Velocity = Total Displacement / Total Time<br /><br />For the first half of the swim, the athlete swims from the north end to the south end of the pool. The total displacement is 50.0 m and the total time is 20.0 s.<br /><br />Average Velocity = 50.0 m / 20.0 s = 2.5 m/s<br /><br />Therefore, the average velocity for the first half of the swim is 2.5 m/s.<br /><br />b. To find the average velocity for the second half of the swim, we can use the same formula:<br /><br />For the second half of the swim, the athlete swims from the south end back to the north end of the pool. The total displacement is -50.0 m (since it's in the opposite direction) and the total time is 22.0 s.<br /><br />Average Velocity = -50.0 m / 22.0 s = -2.27 m/s<br /><br />Therefore, the average velocity for the second half of the swim is -2.27 m/s.<br /><br />c. To find the average velocity for the roundtrip, we can use the same formula:<br /><br />For the roundtrip, the athlete swims from the north end to the south end and back to the north end of the pool. The total displacement is 0 m (since the athlete ends up at the starting position) and the total time is 20.0 s + 22.0 s = 42.0 s.<br /><br />Average Velocity = 0 m / 42.0 s = 0 m/s<br /><br />Therefore, the average velocity for the roundtrip is 0 m/s.<br /><br />4. a. To find how much sooner the faster student arrives at a destination 780 m away, we can use the formula:<br /><br />Time = Distance / Speed<br /><br />For the slower student:<br /><br />Time = 780 m / 0.90 m/s = 866.67 seconds<br /><br />For the faster student:<br /><br />Time = 780 m / 1.90 m/s = 412.11 seconds<br /><br />The difference in time is:<br /><br />Difference = 866.67 seconds - 412.11 seconds = 454.56 seconds<br /><br />Therefore, the faster student arrives 454.56 seconds sooner than the slower student.<br /><br />b. To find how far the students would have to walk so that the faster student arrives 5.50 min before the slower student, we can use the formula:<br /><br />Distance = Speed x Time<br /><br />First, we need to convert the time to seconds:<br /><br />5.50 min = 5.50 x 60 s = 330 s<br /><br />For the slower student:<br /><br />Distance = 0.90 m/s x 330 s = 297 m<br /><br />For the faster student:<br /><br />Distance = 1.90 m/s x 330 s = 627 m<br /><br />The difference in distance is:<br /><br />Difference = 627 m - 297 m = 330 m<br /><br />Therefore, the students would have to walk 330 m so that the faster student arrives 5.50 min before the slower student.