The speed of a falling skydiver is vt 58let3 in metres per

The speed of a falling skydiver is v(t) = 58(l-e^-t/3) in metres per second, where t is the number of seconds since leaving the plane. Then the terminal velocity is L = lim v(t) = The difference |v(t) - L| can be simplified to takes (to the nearest second) for the skydiver 1 So the time it to get within 1/100 metres per second of the terminal velocity is seconds to get within 1/10000 metres per second of the terminal velocity is seconds

Solution

v(t)=58(1-e^-t/3)

lim_t-> v(t)

=lim_t-> 58(1-e^-t/3)

=58(1-0)

=58

|58(1-e^-t/3)- 58|<1/100

|58-58e^-t/3- 58|<1/100

|58e^-t/3|<1/100

|1/e^t/3|<1/5800

|e^t/3|>5800

|t/3|>ln(5800)

|t|>3ln5800

|t|>26

to go with in 26 sec

|58(1-e^-t/3)- 58|<1/10000

|58-58e^-t/3- 58|<1/10000

|58e^-t/3|<1/10000

|1/e^t/3|<1/580000

|e^t/3|>580000

|t/3|>ln(580000)

|t|>3ln580000

|t|>39.8

to go with in 39.8 sec