The velocity v(t)of a particle as a function of time is given by v(t)= (2.3 m/s)+ (4.1 m/s²)t - (6.2 m/s³)t².What is the average acceleration of the particle between t = 1.0 s and t = 2.0 s? A)-13 m/s² B)-15 m/s² C)13 m/s² D)15 m/s² E)0 m/s²

Question

Quizplus · Accepted Answer

The average acceleration of a particle over a time interval is given by the change in velocity over the change in time. First, we need to correct the notation for the velocity function, which is $v(t) = (2.3 \, \text{m/s}) + (4.1 \, \text{m/s}^2)t - (6.2 \, \text{m/s}^3)t^2$. To find the average acceleration between $t = 1.0 \, s$ and $t = 2.0 \, s$, we calculate the velocities at these times and then find the change in velocity over the time interval.At $t = 1.0 \, s$, $v(1) = 2.3 + 4.1(1) - 6.2(1)^2 = 2.3 + 4.1 - 6.2 = 0.2 \, \text{m/s}$.At $t = 2.0 \, s$, $v(2) = 2.3 + 4.1(2) - 6.2(2)^2 = 2.3 + 8.2 - 24.8 = -14.3 \, \text{m/s}$.The change in velocity ($\Delta v$) is $-14.3 \, \text{m/s} - 0.2 \, \text{m/s} = -14.5 \, \text{m/s}$.The change in time ($\Delta t$) is $2.0 \, s - 1.0 \, s = 1.0 \, s$.Thus, the average acceleration ($a_{avg}$) is $\frac{\Delta v}{\Delta t} = \frac{-14.5 \, \text{m/s}}{1.0 \, s} = -14.5 \, \text{m/s}^2$, which is not exactly listed among the options, but the closest match and correct interpretation of the calculation is -15 m/s^2, considering possible rounding or transcription error in the options.

The Velocity V(t)of a Particle as a Function of Time