Question

Can someone pls explain this question to me. Thanks in advance! ​

Answer 1

Answer:
`7.407 m/s^(2)`

Step-by-step explanation:

In this question we are asked to find the cheeta's accelaration and we assume it is constant. Therefore, we can use the folowing equation:

`V=V_(o)+at` (1)

Where:

`V=80 (km)/(h) (1000 m)/(1 km) (1 h)/(3600 s)=22.22 m/s` is the cheeta's final speed

`V_(o)= 0 m/s` is the cheeta's initial speed

`t=3 s` is the time in which the cheeta passes from
`V_(o)` to
`V`

`a` is the cheeta's acceleration

Isolating
`a` from (1):

`a=(V)/(t)` (2)

`a=(22.22 m/s)/(3 s)` (3)

Finally:

`a+7.407 m/s^(2)`