Mean/ SD question on ATV... checking answers?

[IUPAC]

Polaris is interested in the weight of their latest all terrain vehicle (ATV). An ATV consists of 1 body and 4 wheels. Polaris knows that the ATV bodies have a mean weight of 50kg with a variance of 16. ATV wheels have a mean weight of 22kg with a variance of 2.25. The weights of all the parts are independent of each other. The weights of the body as well as the wheels are normally distributed.
(a) What is the mean and standard deviation of the weight of a fully assembled ATV?
So for mean... I did 50kg + 4(22) = 138
for SD, would it be 52? ([16 + 4²(2.25)]
or would I just add the variance 4 times (one for each wheel), then add the mean of the body and square root that? I treated "4 wheels" as being more like multiplying ** a constant...

(c) Polaris estimates that the production cost is $10 per kg of the ATV weight. What are the mean and standard deviation of the production cost of an ATV?
mean = 1380?
SD = 5200?

Thanks in advance.!
Could you explain why it's not 4? Or rather, why you just added the variance of 2.25 4 times instead of treating it as multiplying ** a constant?
Thank you both and Rozeta for replying back
Oh alright. That makes sense... cause it's a constant then x_X... thanks a lot!

[rozeta53]

mean=50kg + 4(22) = 138 kg (right)
SD=[16 + 4(2.25)]=5

(c)
mean = 1380 $ ok
SD = 50 $

EDIT:
It's because:
mean(A+B+C+D+E)=
mean(A)+mean(B)+mean(C)+mean(D)+mean(E)
and
The weights of all the parts are independent of each other ==>
var(A+B+C+D+E)=
=var(A)+var(B)+var(C)+var(D)+var(E)

In your case
mean(B)=mean(C)=mean(D)=mean(E)
var(B)=var(C)=var(D)=var(E)

The weights of the wheels are independent of each other ==>
some could weigh less, some could weigh more.

In case of 4² it would mean that all 4 wheels have the same weight.