Question

In a 100. m race, Marge crosses the finish line after 10.8 s. Accelerating uniformly, Marge took 2.20 s to attain maximum speed, which she maintained for the rest of the race. What was Marge's acceleration?

Answer 1

3.50 m/s^2

Step-by-step explanation:

the race is of 100 m.

time after which Marge crosses the finish line after t= 10.8 sec

time take by her to attain maximum speed t_o= 2.20 sec

a=
`(v)/(t_0)`

we know that

s= ut +1/2at^2

100= v
`100= v*10.8+ 0.5a10.8^2`

put a=
`(v)/(t_0)` here t_o= 2.20 sec

`100=v*10.8+ 0.5 (v)/(2.20) 10.8^2`

v= 7.72 m/s

now Marge acceleration a=
`(7.72)/(2.20)`= 3.50 m/s^2