A skydiver jumps from a stationary hot air balloon and falls vertically before opening her parachute. Her speed at various times after jumping is shown in the table below: Time, t / s 0 1 2 4 6 10 14 Speed, v / m/s 0 9.8 19.4 35 44 50 50 (a) Calculate the skydiver's acceleration…

Answer: ((a)) 9.8 m/s^2, which is equal to g, since air resistance is negligible in the first second of the fall. ((b)) Acceleration between t = 4 s and t = 6 s is 4.5 m/s^2, which is smaller than the 9.8 m/s^2 found in (a). ((c)) Terminal velocity; the resultant force on her becomes zero because air resistance has grown to equal her weight in magnitude, so with zero resultant force there is zero acceleration and her speed stays constant.. Part (a): Acceleration during the first second Acceleration is the change in speed per unit time, a = v/ t. Between t = 0 and t = 1 s: a = 9.8 m/s - 0 m/s/1 s = 9.8 m/s^2 This is equal to g ≈ 9.8 m/s^2, the acceleration of free fall. This…

IGCSE Physics 0625: Free Fall, Acceleration and Terminal Velocity

Time, $t$ / s	$0$	$1$	$2$	$4$	$6$	$10$	$14$
Speed, $v$ / m/s	$0$	$9.8$	$19.4$	$35$	$44$	$50$	$50$

Time,

t

/ s

0

1

2

4

6

10

14

Speed,

v

/ m/s

0

9.8

19.4

35

44

50

50

Part (a): Acceleration during the first second

Acceleration is the change in speed per unit time, $a = \dfrac{\Delta v}{\Delta t}$ . Between $t = 0$ and $t = 1\text{ s}$ :

$a = \frac{9.8\text{ m/s} - 0\text{ m/s}}{1\text{ s}} = 9.8\text{ m/s}^2$

This is equal to $g \approx 9.8\text{ m/s}^2$ , the acceleration of free fall. This makes sense: right at the start of the fall her speed is low, so air resistance acting on her is very small compared with her weight, and she accelerates almost as if there were no air resistance at all.

Part (b): Comparing the acceleration later in the fall

Between $t = 4\text{ s}$ and $t = 6\text{ s}$ , her speed changes from $35\text{ m/s}$ to $44\text{ m/s}$ :

$a = \frac{\Delta v}{\Delta t} = \frac{44\text{ m/s} - 35\text{ m/s}}{6\text{ s} - 4\text{ s}} = \frac{9\text{ m/s}}{2\text{ s}}$

$a = 4.5\text{ m/s}^2$

Comparing this with part (a):

$4.5\text{ m/s}^2 < 9.8\text{ m/s}^2$

so the acceleration between $t = 4\text{ s}$ and $t = 6\text{ s}$ is indeed smaller, as required.

Part (c): Explaining the constant speed after $t = 10\text{ s}$

From $t = 10\text{ s}$ to $t = 14\text{ s}$ , the table shows her speed stays constant at $50\text{ m/s}$ . This constant maximum speed reached during a fall is called terminal velocity.

As the skydiver’s speed increases during the fall, the air resistance (drag) acting upwards on her also increases. Eventually, the air resistance grows until it is equal in magnitude and opposite in direction to her weight. At this point the two forces balance, so the resultant force on her is zero. With no resultant force there is no acceleration, so her speed can no longer increase. It stays constant at this terminal velocity.

Final answers

(a) Acceleration in the first second $=$ $9.8\text{ m/s}^2$ , equal to $g$
(b) Acceleration from $t = 4\text{ s}$ to $t = 6\text{ s}$ $=$ $4.5\text{ m/s}^2$ , which is smaller than $9.8\text{ m/s}^2$
(c) The constant speed reached is called terminal velocity; it occurs when air resistance balances weight, giving zero resultant force and zero acceleration

Worked solution

Part (a): Acceleration during the first second

Part (b): Comparing the acceleration later in the fall

Part (c): Explaining the constant speed after t=10 st = 10\text{ s}t=10 s

Final answers

Part (c): Explaining the constant speed after $t = 10\text{ s}$