Question

Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up for 391 m with an acceleration of +18.9 m/s2. A parachute then opens, slowing the car down with an acceleration of -9.92 m/s2. How fast is the racer moving 332 m after the parachute opens?​

Answer 1

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

$v = \sqrt{2as}$

$v = \sqrt{2*18.9*391}$

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

$V= \sqrt{v^2 + 2aS'}$

$V = \sqrt{121.57^2+ 2 (-9.92)(332)}$

V = 90.51 m/s