Question

In a 100. m race, Lisa crosses the finish line after 11.5 s. Accelerating uniformly, Lisa took 2.00 s to attain maximum speed, which she maintained for the rest of the race. What was Lisa's acceleration?

Answer 1

Answer:

4.76 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{v-0}{2}\\\Rightarrow a=\frac{v}{2}\\\Rightarrow v=2a$

Now, this velocity will be the velocity of the later 11.5-2 = 9.5 seconds with no acceleration so,

Distance = Speed × Time

⇒ D = 2a×9.5 = 19a m

Distance traveled during the first two seconds is

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 2+\frac{1}{2}\times a\times 2^2\\\Rightarrow s=2a\ m$

Total distance = 100 m = Distance traveled in 2 seconds + Distance traveled in the 9.5 seconds

100 = 2a+19a

$\\\Rightarrow 100=21a\\\Rightarrow a=4.76\ m/s^2$

Acceleration during the first two seconds is 4.76 m/s².