I have no idea how to do this. Anyone help?
A cyclist passes a race check point at +5.5m/s and then accelerates ata constant rate of +.55 m/s^2. The cyclist forgot to check in at the check point, and after 20s turns back around to head back. How far did the cylist move from the check point to the point of turning back?
After 20 s, the speed increased from 5.5 m/s to
5.5 m/s + 0.55*20 = 11.5 m/s
The average speed during the 20s was
(11.5 + 5.5)/2 = 8.5 m/s
Multiply that by 20 s for the distance travelled.
To find the distance the cyclist moved from the check point to the point of turning back, we need to calculate the total distance covered during two separate intervals: the time it took the cyclist to reach the checkpoint and the time it took the cyclist to turn back.
Let's break down the problem step by step:
Step 1: Find the distance covered from the start to the checkpoint.
Given:
Initial velocity (u) = 0 m/s (since the cyclist starts from rest)
Final velocity (v) = +5.5 m/s
Time (t) = unknown
We can use the formula:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the acceleration is constant, we can rearrange the formula to solve for time:
t = (v - u) / a
Plugging in the values:
t = (5.5 - 0) / 0.55
t = 10 seconds
Now we know that it took 10 seconds to reach the checkpoint.
Step 2: Find the distance covered from the checkpoint to the point of turning back.
Given:
Initial velocity (u) = 5.5 m/s (since the cyclist maintains the same speed after the checkpoint)
Acceleration (a) = 0.55 m/s^2 (the cyclist accelerates at a constant rate)
Time (t) = 20 seconds (the cyclist turns back after 20 seconds)
We can use the formula:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity
t = time
a = acceleration
Plugging in the values:
s = (5.5 * 20) + (0.5 * 0.55 * (20^2))
Simplifying:
s = 110 + 0.5 * 0.55 * 400
s = 110 + 0.5 * 220
s = 110 + 110
s = 220 meters
Therefore, the cyclist moved a total distance of 220 meters from the checkpoint to the point of turning back.
To find the distance the cyclist moved from the check point to the point of turning back, we can divide the problem into two parts: the distance covered during the initial acceleration phase and the distance covered during the deceleration phase.
First, let's calculate the distance covered during the initial acceleration phase.
1. Calculate the initial velocity (v₀) using the formula: v = v₀ + at
Given: v = +5.5 m/s, a = +0.55 m/s², and t = 20 s (time taken to turn back)
Rearranging the formula: v₀ = v - at
v₀ = 5.5 m/s - (0.55 m/s² × 20 s) [substituting the values]
v₀ = 5.5 m/s - 11 m/s
v₀ = -5.5 m/s
2. Calculate the distance covered during the initial acceleration phase (d₁) using the formula: d = v₀t + (1/2)at²
Given: v₀ = -5.5 m/s, a = +0.55 m/s², and t = 20 s
d₁ = (-5.5 m/s × 20 s) + (0.5 × 0.55 m/s² × (20 s)²) [substituting the values]
d₁ = -110 m + (0.5 × 0.55 m/s² × 400 s²)
d₁ = -110 m + (0.5 × 0.55 m/s² × 40000 s²)
d₁ = -110 m + 11000 m
d₁ = 10890 m
After calculating, we find that the distance covered during the initial acceleration phase (from the check point to the turning point) is 10890 meters.
Now, let's move on to the distance covered during the deceleration phase (returning to the check point).
Since the cyclist is decelerating at the same constant rate, the final velocity would be the same magnitude as the initial velocity, but in the opposite direction.
3. Calculate the distance covered during the deceleration phase (d₂) using the formula: d = v₀t + (1/2)at²
Given: v₀ = -5.5 m/s, a = -0.55 m/s², and t = 20 s
d₂ = (-5.5 m/s × 20 s) + (0.5 × -0.55 m/s² × (20 s)²) [substituting the values]
d₂ = -110 m + (0.5 × -0.55 m/s² × 400 s²)
d₂ = -110 m + (-0.5 × 0.55 m/s² × 400 s²)
d₂ = -110 m + (-0.5 × 0.55 m/s² × 40000 s²)
d₂ = -110 m + (-11000 m)
d₂ = -11110 m
After calculating, we find that the distance covered during the deceleration phase (from the turning point back to the check point) is -11110 meters.
Since distance cannot be negative, we take the magnitude of the displacement.
4. Calculate the magnitude of the displacement by taking the absolute value of d₂:
|d₂| = |-11110 m|
|d₂| = 11110 m
Therefore, the cyclist moved a distance of 11,110 meters from the check point to the point of turning back.