To find the displacement and total distance traveled by the particle with the given velocity function \( v(t) = 3t - 7 \) over the interval \( 0 \leq t \leq 3 \), we proceed with the following steps:
(a) Find the Displacement
Displacement can be found by integrating the velocity function over the given interval:
\[ \text{Displacement} = \int_{0}^{3} v(t) , dt = \int_{0}^{3} (3t - 7) , dt \]
Now, we calculate the integral:
-
Find the antiderivative: \[ \int (3t - 7) , dt = \frac{3t^2}{2} - 7t \]
-
Evaluate the definite integral from \( t = 0 \) to \( t = 3 \): \[ \left[ \frac{3(3)^2}{2} - 7(3) \right] - \left[ \frac{3(0)^2}{2} - 7(0) \right] \]
Simplifying this: \[ \left[ \frac{3 \cdot 9}{2} - 21 \right] - [0] \] \[ = \left[ \frac{27}{2} - 21 \right] \] \[ = \frac{27}{2} - \frac{42}{2} = \frac{-15}{2} \]
Thus, the displacement is: \[ \text{Displacement} = -\frac{15}{2} \text{ meters} = -7.5 \text{ meters} \]
(b) Find the Total Distance Traveled
To find the total distance traveled, we need to determine when the velocity changes sign, indicating a change in direction. We set the velocity function to zero:
\[ 3t - 7 = 0 \implies 3t = 7 \implies t = \frac{7}{3} \approx 2.33 \]
Now, we split the integral at \( t = \frac{7}{3} \) since the velocity is negative when \( t < \frac{7}{3} \) and positive when \( t > \frac{7}{3} \).
-
From \( 0 \) to \( \frac{7}{3} \) (where \( v(t) < 0 \)): \[ \text{Distance}1 = -\int{0}^{\frac{7}{3}} v(t) , dt = -\int_{0}^{\frac{7}{3}} (3t - 7) , dt \] The antiderivative is the same: \[ = -\left[ \frac{3t^2}{2} - 7t \right]_{0}^{\frac{7}{3}}
\] Calculating this gives: \[ = -\left[ \frac{3(\frac{7}{3})^2}{2} - 7\left(\frac{7}{3}\right) \right] \] \[ = -\left[ \frac{3 \cdot \frac{49}{9}}{2} - \frac{49}{3} \right] \] \[ = -\left[ \frac{147}{18} - \frac{294}{18} \right] = -\left[ -\frac{147}{18} \right] = \frac{147}{18} = \frac{49}{6} \] -
From \( \frac{7}{3} \) to \( 3 \) (where \( v(t) > 0 \)): \[ \text{Distance}2 = \int{\frac{7}{3}}^{3} v(t) , dt = \int_{\frac{7}{3}}^{3} (3t - 7) , dt \] This is calculated as follows: \[ = \left[ \frac{3t^2}{2} - 7t \right]_{\frac{7}{3}}^{3}
\] Evaluating this gives: \[ = \left[ \frac{3(3)^2}{2} - 7(3) \right] - \left[ \frac{3(\frac{7}{3})^2}{2} - 7\left(\frac{7}{3}\right) \right] \] The first term is \( \frac{-15}{2} \) (calculated above) and the second term is \( -\frac{49}{6} \).
The total distance traveled is then:
\[ \text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 = \frac{49}{6} + \frac{49}{6} = \frac{98}{6} = \frac{49}{3} \text{ meters} \approx 16.33 \text{ meters} \]
Summary of Results
(a) Displacement: \(-7.5\) meters
(b) Total distance traveled: \(\frac{49}{3} \approx 16.33\) meters
1 answer