Question

A person walking for 0.25 min. can travel 10.5 m and then stops to rest. Find the person’s initial velocity and the person’s acceleration.

Answer 1

Solution1:

0.25 min = 15 secs

a = (v -
`v_(0)`) / t

v = 0 (stops to rest), t = 15

⇒ a = -
`v_(0)`/15

x =
`v_(0)`t +
`(1)/(2)`a
`t^(2)`

⇒ 10.5 =
`v_(0)` * 15 +
`(1)/(2)` *
`(-v0)/(15)` * 225

10.5 =
`v_(0)` * 15 -

10.5 =
`v_(0)` * 7.5

`v_(0)` = 1.4 (m/s)

⇒ a = -
`(7)/(75)` (m/s^2)

Solution2: