Question

A drag racer, starting from rest, speeds up for 402 m with an acceleration of +17.0 m/s2. A parachute then opens, slowing the car down with an acceleration of –6.10 m/s2. How fast is the racer moving 3.50 × 102 m after the parachute opens?

Answer 1

The racer is running at a speed of approximately 96.943 meters per second at 350 meters after the parachute is opened.

According to this question, the drag race accelerates and later decelerates uniformly and we can determine the final speed of the vehicle by the following kinematic formula:

$v = \sqrt{v_{o}^{2}+2\cdot a\cdot s}$ (1)

Where:

• $v_{o}$ - Initial speed, in meters per second.
• $v$ - Final speed, in meters per second.
• $a$ - Acceleration, in meters per square second.
• $s$ - Traveled distance, in meters.

Acceleration phase

If we know that $v_{o} = 0\,\frac{m}{s}$, $a = 17\,\frac{m}{s^{2}}$ and $s = 402\,m$, then the final speed of the drag racer is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2} + 2\cdot \left(17\,\frac{m}{s^{2}} \right)\cdot (402\,m)}$

$v \approx 116.910\,\frac{m}{s}$

Deceleration phase

If we know that $v_{o} \approx 116.910\,\frac{m}{s}$, $a = -6.10\,\frac{m}{s^{2}}$ and $s = 350\,m$, then the final speed of the drag racer is:

$v = \sqrt{\left(116.910\,\frac{m}{s} \right)^{2} + 2\cdot \left(-6.10\,\frac{m}{s^{2}} \right)\cdot (350\,m)}$

$v \approx 96.943\,\frac{m}{s}$

The racer is running at a speed of approximately 96.943 meters per second at 350 meters after the parachute is opened.

We kindly invite you to check this question related to uniform accelerated motion: brainly.com/question/12920060