Question

a sprinter running a 100m dash leaves the starting block and accelerates to a maximum velocity of 11m/s at 6s into the race. the sprinter maintains this velocity for 2s , and then slows down until crossing the finish line 11s after beginning the race. a) what was the sprinter's average acceleration during the first 6s of the race? b) what was the sprinter's average acceleration from 6 to 8 s into the race?

Answer 1

Answer:

Part(a): The average acceleration is $1.83~m~s^{-2}$ during the first 6s.

Part(b): The average acceleration from 6s to 8s is zero.

Explanation:

Part(a):

The average acceleration is defined as the rate at which velocity is changing with respect to time. So if '$v_{i}$', '$v_{f}$' and '$a_{av}$' represents the initial velocity, final velocity, and average acceleration of a particle, then mathematically

$a_{av} = \dfrac{v_{f} - v_{i}}{t}$

where '$t$' is the time taken by the particle to achieve the velocity '$v_{f}$' starting from initial velocity '$v_{i}$'

Given in the problem, $v_{i} = 0, v_{f} = 11~m~s^{-1}~and~t = 6~s$.

So the average acceleration($a_{av}$) during the first 6 s will be

$a_{av} = \dfrac{11 - 0}{6}~m~s^{-2} = 1.83~m~s^{2}$

Part(b):

During the time between 6 s to 8 s, as mentioned in the problem, the sprinter maintains the constant velocity. So the average acceleration during this time interval will be zero.