A go-kart passes a marker post on a straight track with a velocity of 3 m s^-1 and then accelerates uniformly at 1.5 m s^-2. Find the velocity of the go-kart 6 seconds after it passes the marker post.

AS Level Maths 9709: Constant Acceleration. Finding Velocity with v = u + at

The go-kart moves with constant acceleration, so we can use the suvat equation that links velocity, initial velocity, acceleration and time:

$v = u + at$

Here:

$v = 3 + (1.5)(6) = 3 + 9 = 12$

So $v = 12\text{ m s}^{-1}$ .

Working the multiplication separately first: $1.5 \times 6$ . Since $1.5 \times 6 = 1.5\times 5 + 1.5 = 7.5+1.5=9$ , this confirms $at = 9$ .

Adding the initial velocity: $u + at = 3 + 9 = 12$ .

This matches Step 2 exactly, so $v = 12\text{ m s}^{-1}$ .

A ( $9\text{ m s}^{-1}$ ): this is just $at$ , with the initial velocity $u=3$ left out entirely.
C ( $11\text{ m s}^{-1}$ ): comes from miscalculating $1.5\times 6$ as $8$ instead of $9$ , then adding $u=3$ .
D ( $10.5\text{ m s}^{-1}$ ): comes from adding $u+a+t = 3+1.5+6$ directly, instead of substituting into $v=u+at$ .

$\boxed{v = 12\text{ m s}^{-1}} \quad \text{(Option B)}$