A student drives the 105-mi trip back to campus after spring break and travels with an average speed of 53 mi/h for 1 hour and 30 minutes for the first part of the trip. (a) What distance was traveled during this time?

79.5 miles.

Explanation:

From the question given above, the following data were obtained:

Average speed = 53 mi/h

Time = 1 hr 30 minutes

Distance travelled =?

Next, we shall convert 1 hr 30 minutes to hours. This can be obtained as follow:

Recall:

1 hr = 60 mins

Therefore,

1 hr 30 mins = 60 + 30 = 90 minutes.

60 mins = 1 hr

Therefore,

90 mins = 90 mins × 1 hr / 60 mins

90 mins = 1.5 hr

Finally, we shall determine the distance travelled during the time. This can be obtained as follow:

Average speed = 53 mi/h

Time = 1.5 hr

Distance travelled =?

Average speed = Distance travelled / time

53 = Distance travelled / 1.5

Cross multiply

Distance travelled = 53 × 1.5

Distance travelled = 79.5 miles

the ima (higher than 1m) will be greater than the ima with the height of 1m.