Question

-If you were accelerated at the rate of 1ft/sec^2, how fast would you be going at the end of the day, month and year? How would your final speed for each time period compare with the speed of light? Repeat the problem (variation) only with acceleration of 32ft/sec^2.

-Name and state laws and/or principles used

Answer 1

Given,

a = 1 ft/s^2 = 0.305 m/s^2

We know that

v = u + at = 0 + at = at

at the end of the day, t = 24 x 60 x 60 = 86400

v1 = .305 x 86400 = 2.64 x 10^4 m/s

at the end of the month, t = 2.63 x 10^6 s

v2 = .305 x 2.63 x 10^6 = 8.02 x 10^5 m/s

at the end of the year, t = 3.15 x 10^7 s

v3 = 0.305 x 3.15 x 10^7 = 9.61 x 10^6 m/s

c = 3 x 10^8 m/s

v1 = 2.64 x 10^4 c/(3 x 10^8) = 8.8 x 10^-5 c

v2 = 8.02 x 10^5 c/(3 x 10^8) = 2.67 x 10^-3 c

v3 = 9.61 x 10^6 c/(3 x 10^8) = 0.032 c

For this we have used the equation of motion.