Question: A sprinter accelerates uniformly to his top speed after running 30 metres of a 100-metre race. He maintains this speed for the remainder of the race and takes 10.4 seconds to complete it. Find :
a) the top speed of the athlete
b) the time taken to reach the top speed
Working:
$$
a(t)= a
$$
$$
v(t) = \int a(t) \space dt
$$
$$
v(t) = at
$$
$$
r(t) = \int v(t) \space dt
$$
$$
r(t) =\frac{at^2}2
$$
$$
30 =\frac{at^2}2
$$
$$
60 = at^2
$$
$$
t = \sqrt {\frac{60}a}
$$
I have no clue where to go from here, any help appreciated.
Fine so far.
Now (after 30 metres) the sprinter is at speed $v = a t = \sqrt{60 a}$. To complete the following 70 metres until the finish with constant speed, it takes time $t_2$. We then have $70 = v t_2$ and hence $t_2 = 70/v = 70 / \sqrt{60 a}$. So the (given) total time it takes him is $10.4 = t + t_2 = \sqrt {\frac{60}a} + 70 / \sqrt{60 a} = \sqrt {\frac{60}a} (1 + 70/60) $. This gives $$ a = 60 (\frac{1 + 70/60}{10.4})^2 = \frac{125}{48} \simeq 2.604 $$
Now we can answer the two questions:
1) The top speed is $v = \sqrt{60 a} = \sqrt{60 \frac{125}{48}} = 25/2 = 12.5$ metres / second.
2) The time to reach top speed is $t = \sqrt{60 / a} = \sqrt{60 \frac{48}{125}} = 24/5 = 4.8$ seconds.