Question

A sprinter accelerates from rest to 9.00 in 1.38 sec.What is his acceleration in a. m/s2b. m/h2

Answer 1

a) The acceleration of the sprinter is 6.521 meters per square second, b) The acceleration of 84512.16 kilometers per square hour.

Step-by-step explanation:

Note: Statement is incomplete, complete description of the problem is: A sprinter accelerates from rest to
`9.00\,(m)/(s)` in 1.38 seconds. What is his acceleration in a.
`(m)/(s^(2))` and b.
`(km)/(h^(2))`?

a) We assume that sprinter accelerates uniformly, so that acceleration (
`a`), measured in meters per square second, can be obtained from this kinematic expression as function of initial and final velocities (
`v_(o)`,
`v`), measured in meters per second, and time (
`t`), measured in seconds, as well.

`v = v_(o) + a\cdot t`

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

If we know that
`v_(o) = 0\,(m)/(s)`,
`v = 9\,(m)/(s)` and
`t = 1.38\,s`, the acceleration experimented by the sprinter is:

`a = (9\,(m)/(s)-0\,(m)/(s) )/(1.38\,s)`

`a = 6.521\,(m)/(s^(2))`

The acceleration of the sprinter is 6.521 meters per square second.

b) If we know that a hour is equivalent to 3600 seconds and a kilometer is equivalent to 1000 meters, the result of the previous item is converted by using two consecutive unit conversions:

`a = \left(6.521\,(m)/(s^(2)) \right)\cdot \left((1)/(1000)\,(km)/(m) \right)\cdot \left(3600^(2)\,(s^(2))/(h^(2)) \right)`

`a = 84512.16\,(km)/(h^(2))`

The acceleration of 84512.16 kilometers per square hour.