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So I have a problem that reads:

You exert a force of a known magnitude F on a grocery cart of total mass m. The force you exert on the cart points at an angle θ below the horizontal. If the cart starts at rest, determine an expression for the velocity of the cart after it travels a distance d. Ignore friction. Use only F, m d, and $\theta$

The way I thought to do this problem was to use d = 0.5at^2, a = F/m to get:
$$ v = \frac{d}{\sqrt{\frac{2d}{Fcos(\theta)/m}}}$$

because $t=\sqrt{2d/a}$ from d = 0.5at^2

and a in this case would be the cos(theta) times the applied force divided by the mass because a = F/m.

However, this answer is apparently not correct and I am unsure as to why that is.

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Indeed,

(1) $2d = at^2$

For obtaining easily the speed $s$ it is convenient to multiply both sides of this equation with $a$, because in your case,

(2) $s = at$

So, you get

(3) $2ad = s^2$

from which

(4) $s = \sqrt{2ad}$ In (4) you can introduce the expression of $a$, and get

(5) $s = \sqrt{\frac{2F\cos(\theta)d}{m}}$

Your mistake is the $2$, that instead of being in the numerator, appeared somehow in the nominator.

Good luck !

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