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

4.76 m/s²

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(v-0)/(2)\\\Rightarrow a=(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+(1)/(2)at^2\\\Rightarrow s=0* 2+(1)/(2)* a* 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.