A skydiver falls d1 = 495 meters in t1 = 22 s before opening his parachute. After the chute opens, he falls an additional d2 = 470 meters in t2 = 56.5 s. Write an expression that describes the skydiver's average speed, v1 during the period before opening the chute in terms of the given quantities.
Expression :
v1 = ___
d1/t1
hehexd
To calculate the average speed, v1, during the period before opening the parachute, we can use the formula:
Average speed = Total distance / Total time
Since the skydiver falls d1 = 495 meters in t1 = 22 seconds before opening the parachute, the total distance fallen during this period is d1 and the total time taken is t1.
Therefore, the expression for the skydiver's average speed, v1, during the period before opening the chute is:
v1 = d1 / t1
Substituting the given values, the expression becomes:
v1 = 495 meters / 22 seconds
Therefore, the expression for the skydiver's average speed, v1, during the period before opening the chute is:
v1 = 22.5 meters/second
To find the average speed, we need to divide the total distance travelled by the total time taken.
Before opening the parachute, the skydiver falls a distance of d1 = 495 meters in a time of t1 = 22 seconds.
Therefore, we can calculate the average speed, v1, during this period using the formula:
v1 = d1 / t1
Plugging in the given values, the expression for the skydiver's average speed before opening the parachute would be:
v1 = 495 meters / 22 seconds
Simplifying this expression, we get:
v1 = 22.5 meters per second