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CAn object starts from rest, moves in a straight line with a constant acceleration and covers
a distance of 64 m in 4 s. Calculate
(i) its acceleration
(ii) its final velocity
(iii) at what time the object had covered half the total distance
(iv) distance the object had covered in half the total time.

Verified Answer :

GinnyAnswer

Certainly! Let’s solve the problem step-by-step:

### Given:
- Initial velocity (\(\(u\)\)) = 0 m/s (since the object starts from rest)
- Distance covered (\(\(s\)\)) = 64 m
- Time taken (\(\(t\)\)) = 4 s

### (i) Calculate the acceleration (\(\(a\)\))

We use the equation of motion:
\(\[ s = ut + \frac{1}{2}at^2 \]\)

Plugging in the known values:
\(\[ 64 = 0 \cdot 4 + \frac{1}{2}a \cdot 4^2 \]\)
\(\[ 64 = 0 + \frac{1}{2}a \cdot 16 \]\)
\(\[ 64 = 8a \]\)
\(\[ a = \frac{64}{8} \]\)
\(\[ a = 8 \ \text{m/s}^2 \]\)

So, the acceleration is \(\(8 \ \text{m/s}^2\)\).

### (ii) Calculate the final velocity (\(\(v\)\))

We use the equation of motion:
\(\[ v = u + at \]\)

Plugging in the known values:
\(\[ v = 0 + 8 \cdot 4 \]\)
\(\[ v = 32 \ \text{m/s} \]\)

So, the final velocity is \(\(32 \ \text{m/s}\)\).

### (iii) At what time the object had covered half the total distance

Half of the total distance is:
\(\[ \frac{64}{2} = 32 \ \text{m} \]\)

Using the equation of motion:
\(\[ s = ut + \frac{1}{2}at^2 \]\)
Where \(\(s = 32 \ \text{m}\)\), \(\(u = 0\)\), and \(\(a = 8 \ \text{m/s}^2\)\):
\(\[ 32 = 0 + \frac{1}{2} \cdot 8 \cdot t^2 \]\)
\(\[ 32 = 4t^2 \]\)
\(\[ t^2 = \frac{32}{4} \]\)
\(\[ t^2 = 8 \]\)
\(\[ t = \sqrt{8} \]\)
\(\[ t = 2\sqrt{2} \ \text{s} \]\)
\(\[ t \approx 2.83 \ \text{s} \]\)

So, the object covered half the total distance at approximately \(\(2.83 \ \text{s}\)\).

### (iv) Distance the object had covered in half the total time

Half of the total time is:
\(\[ \frac{4}{2} = 2 \ \text{s} \]\)

Using the equation of motion:
\(\[ s = ut + \frac{1}{2}at^2 \]\)
Where \(\(t = 2 \ \text{s}\)\), \(\(u = 0\)\), and \(\(a = 8 \ \text{m/s}^2\)\):
\(\[ s = 0 \cdot 2 + \frac{1}{2} \cdot 8 \cdot 2^2 \]\)
\(\[ s = 0 + \frac{1}{2} \cdot 8 \cdot 4 \]\)
\(\[ s = 4 \cdot 4 \]\)
\(\[ s = 16 \ \text{m} \]\)

So, the distance covered in half the total time is \(\(16 \ \text{m}\)\).

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