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?

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.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft