Kinematics quiz: O-Level Additional Mathematics (SEAB 4049) quiz

By integrating $s$ with respect to $t$

By dividing $s$ by $t$ always

By squaring $s$

By differentiating $s$ with respect to $t$

$a = \dfrac{dv}{dt}$

$a = \displaystyle\int v\,dt$

$a = v \times t$

$a = \dfrac{v}{t}$ always

Differentiate $v$

Integrate $v$ with respect to $t$

Square $v$

Multiply $v$ by acceleration

$s = 0$

$a = 0$

$v = 0$

$t = 0$

The particle is slowing down

The acceleration is negative

The particle has stopped

The particle is moving in the negative direction

$v = 0$

$a = 0$

$s = 0$

$t$ is largest

$2t - 4$

$t^2 - 4$

$2t$

$2t - 4t$

$t = 12$

$t = 2$

$t = 6$

$t = -2$

They are always equal

Displacement is always larger

Distance is always positive and adds movement in every direction; displacement is signed

Distance ignores time

Use the final displacement directly

Take the average velocity times time

Integrate the acceleration once

Find where $v = 0$, compute each leg, and add the magnitudes

$v = 4t + 3$

$v = 4t$

$v = 4t^2 + 3$

$v = 3t + 4$

displacement

velocity

distance

speed