A runner travels 100 m in 10.9 s. Assume that it takes him 1.35 s to reach his top speed, which is then maintained for the rest of the race. Find the following values:
a) acceleration during the first 1.35 s
b) maximum speed
(1/2)*a*(1.35)^2 + Vmax*(9.55) = 100 m
You need a second equation to solve for two unknowns. Here it is:
a*1.35 = Vmax
Therefore
(1/2)*a*(1.35)^2 + 12.89 a = 100
13.80 a = 100
a = 7.246 m/s^2
Vmax = 9.78 m/s
To find the values requested, we will use the formulas of motion. Let's break down the problem into two parts: the initial acceleration and the subsequent constant velocity.
a) Acceleration during the first 1.35 seconds:
Acceleration can be calculated using the equation: acceleration (a) = change in velocity (Δv) / time taken (Δt). In this case, Δv is the change in velocity, and Δt is the time taken.
Given:
Initial velocity (u) = 0 m/s (since the runner starts from rest)
Final velocity (v) = unknown
Time taken (t) = 1.35 s
We need to find Δv, which is the change in velocity during the first 1.35 seconds. We know that:
Δv = v - u
Rearranging the equation:
v = Δv + u
Since the runner starts from rest, the initial velocity (u) is 0 m/s. Therefore:
v = Δv + 0
v = Δv
So, we only need to calculate the change in velocity (Δv). Using the formula:
Δv = a × t
Substituting the given values:
Δv = a × 1.35 s
Now, we need to find the value of Δv using another given information. The runner traveled 100 m in 10.9 seconds. If we only consider the first 1.35 seconds for Δv, we can write the equation:
100 m = Δv × 10.9 s
Rearranging and isolating Δv:
Δv = 100 m / 10.9 s
Therefore, the acceleration during the first 1.35 seconds (a) can be calculated as:
a = Δv / t
a = (100 m / 10.9 s) / 1.35 s
b) Maximum speed:
Since the runner reaches his top speed and then maintains it for the rest of the race, the maximum speed will be maintained throughout the race.
From the given information, the runner traveled 100 m in 10.9 seconds. Therefore, we can calculate the maximum speed (v_max) using the formula:
v_max = distance (d) / time (t)
Substituting the given values:
v_max = 100 m / 10.9 s
So, the maximum speed of the runner is v_max = 100 m / 10.9 s.