# Accelerating sportcare - find time of acceleration

• jgens
In summary, the conversation discussed finding the minimum time required for a sportscar to travel 0.5 mi, given its maximum acceleration and braking rates. The user presented a solution using velocity and time curves and successfully calculated the time required to be 34 s. They also confirmed the correctness of their solution.
jgens
Gold Member

## Homework Statement

A sportscar can accelerate uniformly to 120 mph in 30s. Its maximum breaking rate cannot exceed 0.7g. What is the minimum time required to go 0.5 mi assuming it begins and ends at rest?

N/A

## The Attempt at a Solution

Well, first I argued that the minimum time required to travel 0.5 mi is given by the curve v(t) where the car first accelerates uniformly at its maximum rate until some time t1 and then breaks at its maximum rate until it comes to rest at some time t2. If there were some other velocity v. time curve which allowed the car to travel 0.5 mi in a smaller period of time, say u(t), then for some t' in (0,t2) we must have u(t') > v(t'). Clearly u(t) ≤ v(t) for 0 ≤ t ≤ t1 since the car accelerates at its maximum rate during this period of time. Thus, v(t') < u(t') where t1 < t' < t2. But since the car breaks at its maximum rate during this period for the curve v(t) and v(t2) = 0, this means that u(t2) > 0. Therefore, no such curve u(t) exists.

Now, I just need to calculate the time. To do this, I set up the equations ...

$$x(t_1) = \frac{a_1}{2}t_1^2$$

$$x(t_2)-x(t_1) = (a_1t_1)(t_2-t_1) + \frac{a_2}{2}(t_2-t_1)^2$$

$$a_1t_1 + a_2(t_2-t_1) = 0$$

Where $a_1$ is the initial acceleration and $a_2$ is the breaking rate. Combining the first two equations, and using the substitution $t_2 = t_1(1-a_1a_2^{-1})$, we get ...

$$x(t_2) = t_1^2\left[\frac{-a_1}{2} + a_1\left(1-\frac{a_1}{a_2}\right) + \frac{a_2}{2}\left(1-\frac{a_1}{a_2}\right)^2 - a_2\left(1-\frac{a_1}{a_2}\right) + \frac{a_2}{2}\right]$$

Since $x(t_2) = 1/2$, we can solve this for $t_1$ to get $t_1 = 27 s$ and then use $t_1$ to solve for $t_2$ to get $t_2 = 34 s$.

Does this work out, or have I made a mistake somewhere along the line?

Your solution appears to be correct. You have correctly argued that the minimum time required to travel 0.5 mi is given by the curve v(t), and have used the equations to calculate the time required. Your reasoning is sound and your final answer of 34 s seems to be correct. Good job!

## 1. What is acceleration in sports and why is it important?

Acceleration in sports refers to the rate at which an athlete can increase their speed. It is important because it allows athletes to quickly move from a stationary position to a high speed, giving them an advantage in sports such as sprinting, basketball, and football.

## 2. How do you calculate acceleration in sports?

Acceleration in sports is calculated by dividing the change in an athlete's speed by the time it took to achieve that change. The formula is: acceleration = (final speed - initial speed) / time.

## 3. What are some ways to improve acceleration in sports?

There are several ways to improve acceleration in sports, including strength and power training, plyometric exercises, and proper technique. It is also important to have a strong core and lower body, as well as quick reaction time and coordination.

## 4. How can technology be used to measure acceleration in sports?

Technology such as accelerometers and high-speed cameras can be used to measure acceleration in sports. These tools can provide precise data on an athlete's speed, direction, and acceleration, which can be used for training and performance analysis.

## 5. What are the benefits of accelerating sportcare and finding the time of acceleration?

Accelerating sportcare and finding the time of acceleration can help athletes improve their overall performance and reduce the risk of injuries. It can also help coaches and trainers create more effective training programs tailored to each athlete's needs. Additionally, understanding an athlete's acceleration capabilities can give them a competitive edge in their sport.

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