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A sprinter passes the 10 meter mark as 2 seconds pass since the race started. Find the top speed and acceleration of the sprinter during the interval from when the race started to the time he reached the 2 second mark.

I don't understand why the top speed isn't 10 m/s and why the acceleration isn't 5 m/s/s.

I tried explaining to my science teacher that the average speed doesn't logically make sense to be the final velocity in the given interval, but he said the average velocity is always the final velocity. What is the answer?

I tried using an integral which formed a triangle $\frac{1}{2}at^2$ which was the integral of the line $y=at$ where t was the time. This yielded the exact same answer, but apparently it was "wrong." Apparently this integration formula only works on gravity and falling objects.

mvw
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Saketh Malyala
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    You are right that that acceleration formula works with constant acceleration in general, not just gravity. More generally, $x(t) = v_i t + \frac{1}{2}a t^2$ for initial velocity $v_i$ which in your case I assume is zero. – Carser May 18 '16 at 00:19
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    Yes, it is zero. If it were to have an initial velocity, wouldn't the $v_i t$ just b like stacking a rectangle under the integral, to represent the distance traveled by just that speed? – Saketh Malyala May 18 '16 at 00:21
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    Yes, exactly. I also agree with your result that the acceleration is 5 m/s^2 and the final speed at $t=2$ is 10 m/s. Unless as @mvw has pointed out the acceleration is not constant. – Carser May 18 '16 at 00:22
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    The cruicial bit is linear acceleration. What is that supposed to mean? Is it constant acceration $a(t) = a_0$ or $a(t) = a_0 + a_1 t$? – mvw May 18 '16 at 00:23
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    Linear acceleration refers to the acceleration being constant, and the speed being linearly increasing @mvw – Saketh Malyala May 18 '16 at 00:30

2 Answers2

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Under the assumption of constant acceleration, you are correct that $s=\frac 12at^2$. With $s=10, t=2$ we get $a=5$, then $v=at$ gives a final speed of $10$. Clearly the final speed does not have to be the average speed unless the speed is constant. With a constant acceleration, the average speed will be the speed halfway through in time.

Ross Millikan
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The claim about average and final velocity seems not to be correct. Assuming $a(t) = a = \text{const}$, the following holds:

The average velocity is $$ \bar{v} = \frac{1}{t} \int\limits_0^t v(\tau) d\tau = \frac{1}{t} \int\limits_0^t a \tau \, d\tau = \frac{1}{t} \frac{a}{2}t^2 = \frac{1}{2}at $$ where $\dot{v} = a$ led to $v(t) = v_0 + a t = a t$, because $v_0 = 0$ was assumed.

The final velocity is $$ v(t) = \int\limits_0^t a(\tau) \, d\tau = \int\limits_0^t a_ \, d\tau = a t $$ So $v(t) = 2 \bar{v}$.

mvw
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