I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.
My answer:
Scale Drawing Lengths: . . / .
Actual Court Lengths . . .
Scale Factor: inch corresponds to ( ∙ ) inches, or inches, so the scale factor is .
Let = , represent the scale drawing lengths in inches, and represent the actual court lengths in inches. The -values must be converted from feet to inches.
To find actual length:
= =
() = inches, or feet
To find actual width:
= = ( )
= / ∙ /
= inches, or feet
The actual court measures feet by feet. Yes, the lot is big enough for the court Vincent planned. The court will take up the entire width of the lot.
The software rendering this question really butchered it... I think it's saying the displacement vector for the particle is
Differentiate twice with respect to time to get the velocity and acceleration vectors, respectively:
Then at , the acceleration of the particle is
which most closely resembles the first choice.