Certainly! Let’s solve the problem step-by-step:
### Given:
- Initial velocity ([tex]\(u\)[/tex]) = 0 m/s (since the object starts from rest)
- Distance covered ([tex]\(s\)[/tex]) = 64 m
- Time taken ([tex]\(t\)[/tex]) = 4 s
### (i) Calculate the acceleration ([tex]\(a\)[/tex])
We use the equation of motion:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Plugging in the known values:
[tex]\[ 64 = 0 \cdot 4 + \frac{1}{2}a \cdot 4^2 \][/tex]
[tex]\[ 64 = 0 + \frac{1}{2}a \cdot 16 \][/tex]
[tex]\[ 64 = 8a \][/tex]
[tex]\[ a = \frac{64}{8} \][/tex]
[tex]\[ a = 8 \ \text{m/s}^2 \][/tex]
So, the acceleration is [tex]\(8 \ \text{m/s}^2\)[/tex].
### (ii) Calculate the final velocity ([tex]\(v\)[/tex])
We use the equation of motion:
[tex]\[ v = u + at \][/tex]
Plugging in the known values:
[tex]\[ v = 0 + 8 \cdot 4 \][/tex]
[tex]\[ v = 32 \ \text{m/s} \][/tex]
So, the final velocity is [tex]\(32 \ \text{m/s}\)[/tex].
### (iii) At what time the object had covered half the total distance
Half of the total distance is:
[tex]\[ \frac{64}{2} = 32 \ \text{m} \][/tex]
Using the equation of motion:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Where [tex]\(s = 32 \ \text{m}\)[/tex], [tex]\(u = 0\)[/tex], and [tex]\(a = 8 \ \text{m/s}^2\)[/tex]:
[tex]\[ 32 = 0 + \frac{1}{2} \cdot 8 \cdot t^2 \][/tex]
[tex]\[ 32 = 4t^2 \][/tex]
[tex]\[ t^2 = \frac{32}{4} \][/tex]
[tex]\[ t^2 = 8 \][/tex]
[tex]\[ t = \sqrt{8} \][/tex]
[tex]\[ t = 2\sqrt{2} \ \text{s} \][/tex]
[tex]\[ t \approx 2.83 \ \text{s} \][/tex]
So, the object covered half the total distance at approximately [tex]\(2.83 \ \text{s}\)[/tex].
### (iv) Distance the object had covered in half the total time
Half of the total time is:
[tex]\[ \frac{4}{2} = 2 \ \text{s} \][/tex]
Using the equation of motion:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Where [tex]\(t = 2 \ \text{s}\)[/tex], [tex]\(u = 0\)[/tex], and [tex]\(a = 8 \ \text{m/s}^2\)[/tex]:
[tex]\[ s = 0 \cdot 2 + \frac{1}{2} \cdot 8 \cdot 2^2 \][/tex]
[tex]\[ s = 0 + \frac{1}{2} \cdot 8 \cdot 4 \][/tex]
[tex]\[ s = 4 \cdot 4 \][/tex]
[tex]\[ s = 16 \ \text{m} \][/tex]
So, the distance covered in half the total time is [tex]\(16 \ \text{m}\)[/tex].
