Question

Col. John Stapp led the U.S. Air Force Aero Medical Laboratory's research into the effects of higher accelerations. On Stapp's final sled run, the sled reached a speed of 284.4 m/s (632 mi/h) and then stopped with the aid of water brakes in 1.4 s. Stapp was barely conscious and lost his vision for several days but recovered.
A)Determine his acceleration while stopping.
B)Determine the distance he traveled while stopping

Answer 1

Answer:

A.$a=203.14\ \frac{m}{s^2}$

B.s=397.6 m

Explanation:

Given that

speed  u= 284.4 m/s

time t = 1.4 s

here he want to reduce the velocity from 284.4 m/s to 0 m/s.

So the final speed v= 0 m/s

We know that

v= u + at

So now by putting the values

0 = 284.4 -a x 1.4     (here we take negative sign because this is the case of de acceleration)

$a=203.14\ \frac{m}{s^2}$

So the acceleration  while stopping will be $a=203.14\ \frac{m}{s^2}$.

Lets take distance travel before come top rest is s

We know that

$v^2=u^2-2as$

$0=284.4^2-2\times 203.14\times s$

s=397.6 m

So the distance travel while stopping is 397.6 m.

Answer 2

Answer:

$a) acceleration = -203.14 m/s^2\\\\b) distance = 199.1 m$

Explanation:

Given

initial speed u = 284.4 m/s

final speed v = 0 m/s

duration t = 1.4 s

Solution

a)

Acceleration

$a = \frac{v - u}{t} \\\\a = \frac{0-284.4}{1.4} \\\\a = -203.14 m/s^2$

b)

Distance

$v^2 - u^2 = 2as\\\\0^2 - 284.4^2 = 2 \times (-203.14) \times S\\\\S = 199.1 m$