Cycling distance A cyclist rides down a long straight road with a velocity (in m/min) given by v(t) = 400−20t, for 0≤t≤10, where t is measured in minutes.


b. How far does the cyclist travel in the first 10 min?

Verified step by step guidance

1. Identify the velocity function given: \(v(t) = 400 - 20t\), where \(t\) is in minutes and \(v(t)\) is in meters per minute.
2. Recall that the distance traveled over a time interval can be found by integrating the velocity function over that interval. So, the distance \(D\) traveled from \(t=0\) to \(t=10\) is given by the definite integral \(D = \int_0^{10} v(t) \, dt\).
3. Set up the integral explicitly: \(D = \int_0^{10} (400 - 20t) \, dt\).
4. Integrate the function term-by-term: the integral of \(400\) with respect to \(t\) is \$400t\(, and the integral of \)-20t\( with respect to \)t\( is \)-10t^2$.
5. Evaluate the resulting expression \(400t - 10t^2\) at the upper limit \(t=10\) and subtract the value at the lower limit \(t=0\) to find the total distance traveled.

Key Concepts

Velocity as a Function of Time

Velocity describes the rate of change of position with respect to time. In this problem, velocity is given as a function v(t) = 400 - 20t, which means the cyclist's speed changes linearly over time from 400 m/min to 200 m/min during the first 10 minutes.

Definite Integral for Displacement

The total distance traveled over a time interval can be found by integrating the velocity function over that interval. The definite integral of v(t) from t=0 to t=10 gives the net displacement, representing how far the cyclist has traveled in those 10 minutes.

Limits of Integration and Time Interval

The problem specifies the time interval 0 ≤ t ≤ 10 minutes. Setting the correct limits of integration ensures the calculation covers the entire duration of interest, capturing the cyclist's motion from start to 10 minutes exactly.