A sprinter runs a 100 m dash in 12.0 s. She starts from rest with a constant acceleration ax for 3.0 s and then runs with constant speed for the remainder of the race. What is the value of  ax?

To solve this problem, we will break it down into two phases of the sprinter's motion:

1. **Phase 1 (Acceleration phase):**  
   The sprinter accelerates from rest for \( t_1 = 3.0 \, \text{s} \) with a constant acceleration \( a_x \). The distance covered during this phase will be denoted as \( d_1 \).

2. **Phase 2 (Constant speed phase):**  
   After accelerating, the sprinter runs with constant speed for the remaining time \( t_2 = 12.0 \, \text{s} - 3.0 \, \text{s} = 9.0 \, \text{s} \). The distance covered during this phase will be denoted as \( d_2 \).

The total distance is 100 m, so \( d_1 + d_2 = 100 \, \text{m} \).

---

### Phase 1: Accelerating for 3 seconds
The distance covered during the acceleration phase is given by the equation of motion:
\[
d_1 = \frac{1}{2} a_x t_1^2
\]
The velocity at the end of this phase, \( v_1 \), is:
\[
v_1 = a_x t_1
\]

### Phase 2: Constant speed for 9 seconds
During this phase, the sprinter moves with constant speed \( v_1 \) for \( t_2 = 9.0 \, \text{s} \). The distance covered in this phase is:
\[
d_2 = v_1 t_2 = a_x t_1 t_2
\]

---

### Total distance:
The total distance is the sum of \( d_1 \) and \( d_2 \):
\[
d_1 + d_2 = 100
\]
Substituting the expressions for \( d_1 \) and \( d_2 \):
\[
\frac{1}{2} a_x t_1^2 + a_x t_1 t_2 = 100
\]
Now, substituting \( t_1 = 3.0 \, \text{s} \) and \( t_2 = 9.0 \, \text{s} \):
\[
\frac{1}{2} a_x (3.0)^2 + a_x (3.0)(9.0) = 100
\]
\[
\frac{1}{2} a_x (9) + a_x (27) = 100
\]
\[
4.5 a_x + 27 a_x = 100
\]
\[
31.5 a_x = 100
\]
\[
a_x = \frac{100}{31.5}
\]
\[
a_x \approx 3.17 \, \text{m/s}^2
\]

So, the value of the acceleration \( a_x \) is approximately \( 3.17 \, \text{m/s}^2 \).

---

Would you like more details or have any questions?

Here are some related questions you can explore:
1. What would be the final velocity of the sprinter after 3 seconds?
2. How much distance does the sprinter cover in the first 3 seconds?
3. What would happen if the acceleration phase lasted for a different time?
4. How much time would the sprinter need to finish the race if her acceleration was lower?
5. How does the distance-time graph look for this kind of motion?

**Tip:** Breaking motion into distinct phases helps simplify complex motion problems.

A sprinter runs a 100 m dash in 12.0 s. She starts from rest with a constant acceleration ax for 3.0 s and then runs with constant speed for the remainder of the race. What is the value of ax?

Learn how to calculate the constant acceleration of a sprinter who runs a 100m dash in 12.0 seconds. This detailed step-by-step solution covers kinematic equations, including uniform acceleration and distance formulas. Suitable for high school students in Grades 9-12.

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