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An electron has a constant acceleration of $2.5 \text{ m s}^{-2}$. At a certain instant its velocity is $12.1 \text{ m s}^{-1}.$

(Indicate the direction with the sign of your answer.)

$(a)$ What was its velocity $2.5$ s earlier?

$(b)$ What is its velocity $2.5$ s later?

I'm not sure how to proceed. I know that $a=\dfrac{\Delta v}{\Delta t}$, but is it really a matter of plugging in and dividing?

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    $v(t)=v_0+a t$ as the acceleration is constant, you can take it from here – 2017-02-01

1 Answers 1

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\begin{align*} a &= \frac{v-u}{t} \\ v &= u+at \end{align*}

$(a) \quad $ At $t=+2.5$ s, $$v=+12.1+2.5(+2.5)$$

$(b) \quad $ At $t=-2.5$ s, $$v=+12.1+2.5(-2.5)$$