Calculus Arithmetic (5)

In this article, we will explore three calculus problems related to a car’s velocity function. The velocity of a car at time t is given by the equation v(t) = 3t² + 5t – 2.

1. Calculating the total distance traveled by the car in the time interval [0, 2]:
To find the total distance traveled, we need to integrate the absolute value of the velocity function over the given time interval. Here’s the integration process:

∫[0, 2] |v(t)| dt

= ∫[0, 2] |3t² + 5t – 2| dt

We have to divide the interval [0, 2] into two parts at t = 1, where the velocity function changes sign. One part is [0, 1] and the other is [1, 2]. Solving the integral separately for each part:

∫[0, 2] |3t² + 5t – 2| dt
= ∫[0, 1] -(3t² + 5t – 2) dt + ∫[1, 2] (3t² + 5t – 2) dt

After evaluating the integrals, we find:

= [-t³/3 – 5t²/2 + 2t] from 0 to 1 + [t³/3 + 5t²/2 – 2t] from 1 to 2

Plugging in the values:

= {[-1/3 – 5/2 + 2] – [0]} + {[8/3 + 20/2 – 4] – [-1/3 – 5/2 + 2]}
= {-1/3 – 5/2 + 2} + {8/3 + 20/2 – 4 + 1/3 + 5/2 – 2}

Simplifying the terms:

= 1/3 – 5/2
= -13/6

Therefore, the total distance traveled by the car in the time interval [0, 2] is -13/6.

2. Calculating the car’s acceleration at t = 2:
To find the acceleration, we need to calculate the derivative of the velocity function with respect to time. Differentiating v(t) = 3t² + 5t – 2:

a(t) = d(v(t))/dt
= d(3t² + 5t – 2)/dt

Taking the derivative:

= 6t + 5

When t = 2:

a(2) = 6(2) + 5
= 17

Therefore, the car’s acceleration at t = 2 is 17.

3. Calculating the total travel time within the time interval [0, 2]:
The total travel time can be calculated by integrating the derivative of the position function. However, since we are not given the position function explicitly, we cannot calculate it directly. We can only calculate the total distance traveled as mentioned in question 1.

In conclusion, the total distance traveled by the car in the time interval [0, 2] is -13/6, the car’s acceleration at t = 2 is 17, and we cannot determine the total travel time within [0, 2] without additional information about the position function.
微積分算術（5）

在本文中，我們將探討與汽車速度函數相關的三個微積分問題。汽車在時間t的速度由方程式v(t) = 3t² + 5t – 2給出。

1. 計算汽車在時間區間[0, 2]內行駛的總距離：
為了找到行駛的總距離，我們需要對給定的時間區間內速度函數的絕對值進行積分。以下是積分的過程：

∫[0, 2] |v(t)| dt

= ∫[0, 2] |3t² + 5t – 2| dt

我們必須在t = 1處將區間[0, 2]劃分為兩個部分，因為速度函數在這裡改變符號。一部分是[0, 1]，另一部分是[1, 2]。分別解決每個部分的積分：

∫[0, 2] |3t² + 5t – 2| dt
= ∫[0, 1] -(3t² + 5t – 2) dt + ∫[1, 2] (3t² + 5t – 2) dt

計算積分後，我們得到：

= [-t³/3 – 5t²/2 + 2t] 從 0 到 1 + [t³/3 + 5t²/2 – 2t] 從 1 到 2

代入數值：

= {[-1/3 – 5/2 + 2] – [0]} + {[8/3 + 20/2 – 4] – [-1/3 – 5/2 + 2]}
= {-1/3 – 5/2 + 2} + {8/3 + 20/2 – 4 + 1/3 + 5/2 – 2}

簡化數字：

= 1/3 – 5/2
= -13/6