Question

The velocity of a bicycle is given by v(t) = 7t feet per second, where t is the number of seconds after the bike starts moving. How far does the bicycle travel in 3 seconds?

Answer 1

To find the distance the bicycle travels in 3 seconds, integrate the velocity function v(t) = 7t from 0 to 3 seconds to get a total distance of 31.5 feet.

Step-by-step explanation:

To calculate the distance the bicycle travels in 3 seconds, we need to integrate the velocity function v(t) = 7t with respect to time over the interval from 0 to 3 seconds. The distance covered, or displacement, is the integral of velocity v(t) with respect to time t:

1. Write the integral for the distance: ∫ v(t) dt = ∫ 7t dt.
2. Integrate 7t with respect to t over the interval from 0 to 3 seconds: ∫_{0}^{3} 7t dt = [\frac{7}{2}t^2]_{0}^{3}.
3. Calculate the definite integral: [\frac{7}{2}t^2]_{0}^{3} = \frac{7}{2}(3^2) - \frac{7}{2}(0^2) = \frac{7}{2}(9) = \frac{63}{2}.
4. Convert the fraction to a decimal: \frac{63}{2} = 31.5.

Therefore, the bicycle travels a distance of 31.5 feet in 3 seconds.