A young man driving a car needs 100 s to cover a distance of 1000 m. If he accelerates uniformly and starts from rest, calculate: a) the velocity he will have at the end of that distance, b) the velocity he carried for the first quarter of the total distance, c) the time taken to cover the second half of the distance.

We are given the following information:

- Time to cover the total distance \( t = 100 \, \text{s} \)
- Distance covered \( d = 1000 \, \text{m} \)
- The car starts from rest, so the initial velocity \( u = 0 \, \text{m/s} \)
- The acceleration is uniform.

Let's break down the problem step by step to find the answers:

### (a) Final velocity after covering the total distance:

To calculate the final velocity at the end of the distance, we can use the second equation of motion:

\[
d = u t + \frac{1}{2} a t^2
\]

Since \( u = 0 \), the equation simplifies to:

\[
d = \frac{1}{2} a t^2
\]

Substitute the given values:

\[
1000 = \frac{1}{2} a (100)^2
\]

Solving for \( a \):

\[
1000 = 50 a \times 100
\]
\[
a = \frac{1000}{5000} = 0.2 \, \text{m/s}^2
\]

Now, we can calculate the final velocity using the first equation of motion:

\[
v = u + at
\]

Since \( u = 0 \):

\[
v = 0 + 0.2 \times 100 = 20 \, \text{m/s}
\]

### Final velocity at the end of the distance is \( 20 \, \text{m/s} \).

---

### (b) Velocity after the first quarter of the total distance:

The first quarter of the distance is:

\[
\text{First quarter distance} = \frac{1000}{4} = 250 \, \text{m}
\]

We can use the third equation of motion to find the velocity at this point:

\[
v^2 = u^2 + 2ad
\]

Again, since \( u = 0 \):

\[
v^2 = 2 \times 0.2 \times 250
\]
\[
v^2 = 100
\]
\[
v = \sqrt{100} = 10 \, \text{m/s}
\]

### The velocity after covering the first quarter of the distance is \( 10 \, \text{m/s} \).

---

### (c) Time to cover the second half of the distance:

The second half of the total distance is:

\[
\text{Second half distance} = \frac{1000}{2} = 500 \, \text{m}
\]

Now, to find the time for the second half, we need to calculate the time at which the first 500 m is covered. Again, using the second equation of motion:

\[
d = \frac{1}{2} a t^2
\]

For the first half distance:

\[
500 = \frac{1}{2} \times 0.2 \times t^2
\]
\[
500 = 0.1 \times t^2
\]
\[
t^2 = \frac{500}{0.1} = 5000
\]
\[
t = \sqrt{5000} \approx 70.71 \, \text{s}
\]

The time to cover the second half is the total time minus the time to cover the first half:

\[
t_{\text{second half}} = 100 - 70.71 \approx 29.29 \, \text{s}
\]

### The time to cover the second half of the distance is approximately \( 29.29 \, \text{s} \).

---

### Summary of Results:
1. Final velocity at the end of the distance is \( 20 \, \text{m/s} \).
2. Velocity after covering the first quarter of the distance is \( 10 \, \text{m/s} \).
3. Time taken to cover the second half of the distance is \( 29.29 \, \text{s} \).

Let me know if you'd like more details or further clarifications.

---

Here are five related questions:
1. How would the problem change if the initial velocity were not zero?
2. What would the acceleration be if the time was halved but the distance remained the same?
3. How does uniform acceleration differ from non-uniform acceleration in terms of calculating velocities?
4. How can you determine the average velocity over the entire journey?
5. What is the significance of the third equation of motion in solving uniformly accelerated motion problems?

**Tip:** Always make sure you use consistent units (e.g., meters, seconds) when working with motion equations to avoid errors.

Solve a kinematics problem involving uniform acceleration. Learn how to calculate the final velocity after covering 1000 m in 100 s, the velocity after covering the first quarter of the distance, and the time taken for the second half. This problem uses the equations of motion and is suitable for students in Grades 9-12.

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