Ask Question
28 April, 03:24

The rocket-driven sled Sonic Wind No. 2, used for investigating the physiological effects of large accelerations, runs on a straight, level track 1070 m (3500 ft) long. Starting from rest, it can reach a speed of 224 m/s (500 mi/h) in 0.900 s. (a) Compute the acceleration in m/s2, assuming that it is constant. (b) What is the ratio of this acceleration to that of a freely falling body (g) ? (c) What distance is covered in 0.900 s? (d) A magazine article states that at the end of a certain run, the speed of the sled de-creased from 283 m/s (632 mi/h) to zero in 1.40 s and that during this time the magnitude of the acceleration was greater than 40 g. Are these figures consistent?

+3
Answers (1)
  1. 28 April, 04:55
    0
    a) The acceleration of the rocket is 249 m/s².

    b) The acceleration of the rocket is 25 times the acceleration of a free-falling body (25 g),

    c) The distance traveled in 0.900 s was 101 m.

    d) The figures are not consistent. The acceleration of the rocket was 20 g.

    Explanation:

    Hi there!

    a) To calculate the acceleration of the rocket let's use the equation of velocity of the rocket:

    v = v0 + a · t

    Where:

    v = velocity of the rocket.

    v0 = initial velocity.

    a = acceleration.

    t = time.

    We know that at t = 0.900 s, v = 224 m/s. The initial velocity, v0, is zero because the rocket starts from rest.

    v = v0 + a · t

    Solving for a:

    (v - v0) / t = a

    224 m/s / 0.900 s = a

    a = 249 m/s²

    The acceleration of the rocket is 249 m/s²

    b) The acceleration of gravity is ≅ 10 m/s². The ratio of the acceleration of the rocket to the acceleration of gravity will be:

    249 m/s² / 10 m/s² = 25

    So, the acceleration of the rocket is 25 times the acceleration of gravity or 25 g.

    c) The equation of traveled distance is the following:

    x = x0 + v0 · t + 1/2 · a · t²

    Where:

    x = position of the rocket at time t.

    x0 = initial position.

    v0 = initial velocity.

    t = time.

    a = acceleration.

    Since x0 and v0 are equal to zero, then, the equation of position gets reduced to:

    x = 1/2 · a · t²

    x = 1/2 · 249 m/s² · (0.900 s) ²

    x = 101 m

    The distance traveled in 0.900 s was 101 m.

    d) Now, using the equation of velocity, let's calculate the acceleration. We know that at 1.40 s the velocity of the rocket is zero and that the initial velocity is 283 m/s.

    v = v0 + a · t

    0 m/s = 283 m/s + a · 1.40 s

    -283 m/s / 1.40 s = a

    a = - 202 m/s²

    The figures are not consistent because 40 g is equal to an acceleration of 400 m/s² and the magnitude of the acceleration of the rocket was ≅20 g.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “The rocket-driven sled Sonic Wind No. 2, used for investigating the physiological effects of large accelerations, runs on a straight, level ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers