A drag racer, starting from rest, speeds up for 402 m with an acceleration of `+17.0 m//s^(2)`. A parachute then opens, slowing the car down with an acceleration of `-6.10 m//s^(2)`. How fast is the racer moving `3.50xx10^(2)` m after the parachute opens ?

Options

A. 96.9 m/s
B. 65.4 m/s
C. 82.9 m/s
D. 20.1 m/s

Answer

To solve the problem step by step, we will break it down into two parts: the motion from point A to point B (acceleration phase) and the motion from point B to point C (deceleration phase).

### Step 1: Calculate the final velocity at point B

1. **Given Data**:
- Initial velocity (u) = 0 m/s (starting from rest)
- Acceleration (a) = +17.0 m/s²
- Distance (s) = 402 m

2. **Using the equation of motion**:
\[ v^2 = u^2 + 2as \]
Substituting the values:
\[ v_B^2 = 0 + 2 \times 17.0 \times 402 \]
\[ v_B^2 = 13668 \]
\[ v_B = \sqrt{13668} \approx 116.91 \text{ m/s} \]

### Step 2: Calculate the velocity at point C after the parachute opens

1. **Given Data for the second part**:
- Initial velocity at point B (u_B) = 116.91 m/s
- Acceleration (a) = -6.10 m/s² (deceleration)
- Distance (s) = 3.5 x 10² m = 350 m

2. **Using the equation of motion again**:
\[ v_C^2 = u_B^2 + 2as \]
Substituting the values:
\[ v_C^2 = (116.91)^2 + 2 \times (-6.10) \times 350 \]
\[ v_C^2 = 13668 - 4270 \]
\[ v_C^2 = 9398 \]
\[ v_C = \sqrt{9398} \approx 96.94 \text{ m/s} \]

### Final Answer
The speed of the racer moving 350 m after the parachute opens is approximately **96.94 m/s**.

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