A powerboat, starting from rest, maintains a constant acceleration

[hey123a]

Homework Statement

A powerboat, starting from rest, maintains a constant acceleration. After a certain time t, its displacement and velocity are r and v. At time 2t, what would be its displacement and velocity, assuming the acceleration remains the same?
a) 2r and 2v
b) 2r and 4v
c) 4r and 2v
d) 4r and 4v

the correct answer is c

Homework Equations

v = vo + at
x = vo + 1/2at^2

The Attempt at a Solution

r = vo + 1/2at^2
r = (0) + 1/2a(2T)^2
r = 1/2a4T^2

v = vo + at
v = (o) + a(2T)
v = a2T

i don't know what to do next because acceleration is not given so how i could i even isolate anything?

 

[voko]

What does "starting from rest" mean to you?

 

[hey123a]

What does "starting from rest" mean to you?

starting from rest means initial velocity is equal to zero

 

[voko]

So you got $v_1 = at$ and $r_1 = at^2/2$. You have further found at $2t$, $v_2 = a(2t)$ and $r_2 = a(4t^2)/2$. All that you need to do is express $v_2$ via $v_1$ and $r_2$ via $r_1$.

 

[hey123a]

So you got $v_1 = at$ and $r_1 = at^2/2$. You have further found at $2t$, $v_2 = a(2t)$ and $r_2 = a(4t^2)/2$. All that you need to do is express $v_2$ via $v_1$ and $r_2$ via $r_1$.

what do you mean by via

 

[voko]

If you have a = 5b, and c = 10b, then you can express c via a as follows: c = 2a.

 

[hey123a]

If you have a = 5b, and c = 10b, then you can express c via a as follows: c = 2a.

Ah okay so
if v1 = at
and v2 = 2at
then v2 = 2v1

and if r1 = at^2/2
and r2 = a4t^2/2
r2 could be simplified into = (a2t^2)/1
then r2 = 4r1

right? to check,
4r1 = 4(at^2)/2
= 4at^2/2
= (2at^2)/1

which leaves me with 4v and 2v as my answer, which is c which is the correct answer :) thank you