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

Answer:

3.50 m/s^2

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= $\frac{v}{t_0}$

we know that

s= ut +1/2at^2

100= v$100= v\times10.8+ 0.5a10.8^2$

put a= $\frac{v}{t_0}$ here t_o= 2.20 sec

$100=v\times10.8+ 0.5 \frac{v}{2.20} 10.8^2$

v= 7.72 m/s

now Marge acceleration a= $\frac{7.72}{2.20}$= 3.50 m/s^2