To calculate the distance traveled by the particle, we need to find the displacement function by integrating the velocity function.
The displacement function is the integral of the velocity function:
𝑊(𝑡) = ∫ (3𝑡^2 − 2𝑡 + 5) 𝑑𝑡
Evaluating the integral gives us:
𝑊(𝑡) = 𝑡^3 − 𝑡^2 + 5𝑡 + 𝐶
To find the constant of integration, we use the initial condition 𝑊(2) = 0, since the particle starts at t = 2 seconds.
0 = (2)^3 − (2)^2 + 5(2) + 𝐶
0 = 8 - 4 + 10 + 𝐶
𝐶 = -14
So, the displacement function becomes:
𝑊(𝑡) = 𝑡^3 − 𝑡^2 + 5𝑡 - 14
To find the distance traveled, we need to find the absolute value of the displacement between t = 2 seconds and t = 6 seconds:
Distance = |𝑊(6) - 𝑊(2)|
Distance = |(6)^3 - (6)^2 + 5(6) - 14 - [(2)^3 - (2)^2 + 5(2) - 14]|
Distance = |216 - 36 + 30 - 14 - 8 + 4 - 10 + 14|
Distance = |206|
Therefore, the distance traveled by the particle between t = 2 seconds and t = 6 seconds is 206 meters.
The velocity V m/s of a particle in motion is given by 𝑉 = 3𝑡
2 − 2𝑡 + 5.Calculate the distance
travelled by the particle between t= 2 seconds and t = 6 seconds. (3 marks)
1 answer