Answer :
To answer the questions, we will use integration techniques to analyze the given velocity function [tex]\( v(t) = t^2 - 4 \)[/tex].
### Part (a): Net Displacement
The net displacement is found by integrating the velocity function over the given time interval [tex]\([0, 4]\)[/tex].
[tex]\[ \text{Displacement} = \int_{0}^{4} v(t) \, dt = \int_{0}^{4} (t^2 - 4) \, dt \][/tex]
We will integrate each term separately:
[tex]\[ \int_{0}^{4} (t^2 - 4) \, dt = \int_{0}^{4} t^2 \, dt - \int_{0}^{4} 4 \, dt \][/tex]
The first integral is:
[tex]\[ \int_{0}^{4} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{4} = \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3} \][/tex]
The second integral is:
[tex]\[ \int_{0}^{4} 4 \, dt = 4 \int_{0}^{4} 1 \, dt = 4 [t]_{0}^{4} = 4(4 - 0) = 16 \][/tex]
Combining these results, we have:
[tex]\[ \text{Displacement} = \frac{64}{3} - 16 = \frac{64}{3} - \frac{48}{3} = \frac{16}{3} \][/tex]
Therefore, the net displacement of the particle between [tex]\( t = 0 \)[/tex] and [tex]\( t = 4 \)[/tex] is [tex]\( \frac{16}{3} \)[/tex] meters.
### Part (b): Distance Travelled
To find the distance travelled, we need to account for the fact that the particle may change direction, which occurs when the velocity changes sign. We consider the absolute value of the velocity function and integrate it over the given interval.
The velocity function [tex]\( v(t) = t^2 - 4 \)[/tex] changes sign where [tex]\( t^2 - 4 = 0 \)[/tex]:
[tex]\[ t^2 - 4 = 0 \implies t^2 = 4 \implies t = \pm 2 \][/tex]
In the interval [tex]\([0, 4]\)[/tex], the relevant point is [tex]\( t = 2 \)[/tex].
We split the integral at [tex]\( t = 2 \)[/tex] and consider the absolute value of the velocity:
[tex]\[ \text{Distance} = \int_{0}^{2} |t^2 - 4| \, dt + \int_{2}^{4} |t^2 - 4| \, dt \][/tex]
Since [tex]\( t^2 - 4 \)[/tex] is negative for [tex]\( 0 \leq t < 2 \)[/tex] and positive for [tex]\( 2 \leq t \leq 4 \)[/tex], we rewrite the absolute values accordingly:
[tex]\[ \text{Distance} = \int_{0}^{2} (4 - t^2) \, dt + \int_{2}^{4} (t^2 - 4) \, dt \][/tex]
First integral:
[tex]\[ \int_{0}^{2} (4 - t^2) \, dt = \int_{0}^{2} 4 \, dt - \int_{0}^{2} t^2 \, dt \][/tex]
Solving the integrals separately:
[tex]\[ \int_{0}^{2} 4 \, dt = 4 [t]_{0}^{2} = 4(2 - 0) = 8 \][/tex]
[tex]\[ \int_{0}^{2} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \][/tex]
Combining these, we get:
[tex]\[ \int_{0}^{2} (4 - t^2) \, dt = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \][/tex]
Second integral:
[tex]\[ \int_{2}^{4} (t^2 - 4) \, dt = \int_{2}^{4} t^2 \, dt - \int_{2}^{4} 4 \, dt \][/tex]
Solving the integrals separately:
[tex]\[ \int_{2}^{4} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{2}^{4} = \frac{4^3}{3} - \frac{2^3}{3} = \frac{64}{3} - \frac{8}{3} = \frac{56}{3} \][/tex]
[tex]\[ \int_{2}^{4} 4 \, dt = 4 [t]_{2}^{4} = 4(4 - 2) = 8 \][/tex]
Combining these, we get:
[tex]\[ \int_{2}^{4} (t^2 - 4) \, dt = \frac{56}{3} - 8 = \frac{56}{3} - \frac{24}{3} = \frac{32}{3} \][/tex]
Adding the two integrals together:
[tex]\[ \text{Distance} = \int_{0}^{2} (4 - t^2) \, dt + \int_{2}^{4} (t^2 - 4) \, dt = \frac{16}{3} + \frac{32}{3} = \frac{48}{3} = 16 \][/tex]
Therefore, the distance travelled by the particle between [tex]\( t = 0 \)[/tex] and [tex]\( t = 4 \)[/tex] is 16 meters.
### Part (a): Net Displacement
The net displacement is found by integrating the velocity function over the given time interval [tex]\([0, 4]\)[/tex].
[tex]\[ \text{Displacement} = \int_{0}^{4} v(t) \, dt = \int_{0}^{4} (t^2 - 4) \, dt \][/tex]
We will integrate each term separately:
[tex]\[ \int_{0}^{4} (t^2 - 4) \, dt = \int_{0}^{4} t^2 \, dt - \int_{0}^{4} 4 \, dt \][/tex]
The first integral is:
[tex]\[ \int_{0}^{4} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{4} = \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3} \][/tex]
The second integral is:
[tex]\[ \int_{0}^{4} 4 \, dt = 4 \int_{0}^{4} 1 \, dt = 4 [t]_{0}^{4} = 4(4 - 0) = 16 \][/tex]
Combining these results, we have:
[tex]\[ \text{Displacement} = \frac{64}{3} - 16 = \frac{64}{3} - \frac{48}{3} = \frac{16}{3} \][/tex]
Therefore, the net displacement of the particle between [tex]\( t = 0 \)[/tex] and [tex]\( t = 4 \)[/tex] is [tex]\( \frac{16}{3} \)[/tex] meters.
### Part (b): Distance Travelled
To find the distance travelled, we need to account for the fact that the particle may change direction, which occurs when the velocity changes sign. We consider the absolute value of the velocity function and integrate it over the given interval.
The velocity function [tex]\( v(t) = t^2 - 4 \)[/tex] changes sign where [tex]\( t^2 - 4 = 0 \)[/tex]:
[tex]\[ t^2 - 4 = 0 \implies t^2 = 4 \implies t = \pm 2 \][/tex]
In the interval [tex]\([0, 4]\)[/tex], the relevant point is [tex]\( t = 2 \)[/tex].
We split the integral at [tex]\( t = 2 \)[/tex] and consider the absolute value of the velocity:
[tex]\[ \text{Distance} = \int_{0}^{2} |t^2 - 4| \, dt + \int_{2}^{4} |t^2 - 4| \, dt \][/tex]
Since [tex]\( t^2 - 4 \)[/tex] is negative for [tex]\( 0 \leq t < 2 \)[/tex] and positive for [tex]\( 2 \leq t \leq 4 \)[/tex], we rewrite the absolute values accordingly:
[tex]\[ \text{Distance} = \int_{0}^{2} (4 - t^2) \, dt + \int_{2}^{4} (t^2 - 4) \, dt \][/tex]
First integral:
[tex]\[ \int_{0}^{2} (4 - t^2) \, dt = \int_{0}^{2} 4 \, dt - \int_{0}^{2} t^2 \, dt \][/tex]
Solving the integrals separately:
[tex]\[ \int_{0}^{2} 4 \, dt = 4 [t]_{0}^{2} = 4(2 - 0) = 8 \][/tex]
[tex]\[ \int_{0}^{2} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \][/tex]
Combining these, we get:
[tex]\[ \int_{0}^{2} (4 - t^2) \, dt = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \][/tex]
Second integral:
[tex]\[ \int_{2}^{4} (t^2 - 4) \, dt = \int_{2}^{4} t^2 \, dt - \int_{2}^{4} 4 \, dt \][/tex]
Solving the integrals separately:
[tex]\[ \int_{2}^{4} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{2}^{4} = \frac{4^3}{3} - \frac{2^3}{3} = \frac{64}{3} - \frac{8}{3} = \frac{56}{3} \][/tex]
[tex]\[ \int_{2}^{4} 4 \, dt = 4 [t]_{2}^{4} = 4(4 - 2) = 8 \][/tex]
Combining these, we get:
[tex]\[ \int_{2}^{4} (t^2 - 4) \, dt = \frac{56}{3} - 8 = \frac{56}{3} - \frac{24}{3} = \frac{32}{3} \][/tex]
Adding the two integrals together:
[tex]\[ \text{Distance} = \int_{0}^{2} (4 - t^2) \, dt + \int_{2}^{4} (t^2 - 4) \, dt = \frac{16}{3} + \frac{32}{3} = \frac{48}{3} = 16 \][/tex]
Therefore, the distance travelled by the particle between [tex]\( t = 0 \)[/tex] and [tex]\( t = 4 \)[/tex] is 16 meters.
