After jumping from a plane a skydiver opens his parachute and slows to his terminal velocity of 11.5 ms-1 in 8.3 s. His deceleration was -14 m s-². How far had he descended in this time? Give your answer to two significant figures.

Question

After jumping from a plane a skydiver opens his parachute and slows to his terminal velocity of 11.5 ms-1 in 8.3 s. His deceleration was -14 m s-². How far had he descended in this time? Give your answer to two significant figures.

BlackTom AI · Accepted Answer

Use the SUVAT equation s = ut + 1/2 a t^2 (constant acceleration kinematics).

1) Relevant concepts:
- s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time.
- With constant acceleration we can use v = u + at to find u if v, a and t are known.
- Average velocity method s = (u + v)/2 · t is an equivalent check when acceleration is constant.

2) Step-by-step reasoning and intermediate calculations:
- Given: v = 11.5 m s^-1 (final downward speed), a = -14 m s^-2 (deceleration, opposite to downward direction), t = 8.3 s.
- Find initial velocity u from v = u + a t:
  u = v - a t = 11.5 - (-14)(8.3) = 11.5 + 116.2 = 127.7 m s^-1.
- Now compute displacement using s = u t + 1/2 a t^2:
  t^2 = 8.3^2 = 68.89 s^2.
  u t = 127.7 × 8.3 = 1059.91 m.
  (1/2) a t^2 = 0.5 × (-14) × 68.89 = -7 × 68.89 = -482.23 m.
  So s = 1059.91 - 482.23 = 577.68 m descended.

3) Rounding to two significant figures:
- 577.68 rounded to two significant figures → 5.8 × 10^2 m (commonly written as 580 m, but in two significant-figure scientific form 5.8e2).

4) Why this is correct:
- The calculation correctly uses kinematic relations for constant acceleration, respects sign of deceleration, and carries full intermediate precision before final rounding.

5) Common incorrect approaches:
- Using v as u in s = ut + 1/2 a t^2 would give an incorrect (and negative) result.
- Forgetting to include the sign of a or rounding too early can produce errors.

Conclusion: The skydiver descended 5.8 × 10^2 m in the 8.3 s, which matches the provided answer 5.8e2.

After jumping from a plane a skydiver opens his parachute and slows to his terminal velocity of 11.5 ms-1 in 8.3 s. His deceleration was -14 m s-². How far had he descended in this time? Give your answer to two significant figures.Numerical

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