Question

A rocket-driven sled running on a straight level track is used to investigate the physiological effects of large accelerations on humans. One such sled can attain a speed of 1000 rniles/hr in 1.8 sec starting from rest. (1 mile = 5280 ft) a. Assume the acceleration is constant and compare it to g. (g=9.8 m/s2 = 32 ft/ s2) b. What is the distance traveled in this time?

Answer 1

Answer:

A. a = 814.815 ft/s^2.

B. Distance, S = 320 ft.

Explanation:

Equations of motion

i. vf = vi + a*t

ii. S = vi*t + 1/2*(a*t)

iii. vf^2 = vi^2 + 2a*S

Given:

vi = 0 ft/s

vf = 1000 miles/hr

t = 1.8 s

g = 32 ft/ s2

Converting miles/hr to ft/s,

1 mile = 5280 ft

Also, 1 hr = 60 mins * 60s

= 3600 s

Therefore, 1000 miles/hr * 5280 ft/1 mile * 1 hr/3600 s

= 1466.7 ft/s.

A.

Using the i. Equation of motion,

1466.7 = 0 + a*1.8

a = 814.815 ft/s^2

Comparing a to g,

a = (814.815/32) * g

= 25.46 g

B.

Using the ii. Equation of motion,

S = 1/2 * (814.815) * (1.8)^2

= 1320 ft.