Given the equation Distance = 25.96884t^1.54177, with the instructions (CAREFUL! THIS IS THE POSITION OF THE CAR AT T-SECONDS … NOT ITS VELOCITY.) How would I proceed to solve Step C in my homework?
C. The slip claims that his speed at the 1/8 mile mark is 83.85 mph. Determine if that is accurate using the position equation in part B.
C. The slip claims that his speed at the 1/8 mile mark is 83.85 mph. Determine if that is accurate using the position equation in part B.
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x = 25.96884t^1.54177
Speed v is dx/dt = (25.96884*1.54177) t^0.54177 ~ 40.0379784468 t^0.54177
but we do not yet know time t at the 1/8 mile mark.
Assuming distance x is based on feet, 1/8 mile = 5280/8 = 660 ft
so solve 25.96884t^1.54177 = 660 (on calculator) to find t
t = 8.1536 secs
v = 40.0379784468*(8.1536)^0.54177 ~124.7998 ft/sec
15 mph = (15*5280)/(3600) = 22 ft/sec
v = 124.7998 *(15/22) ~ 85.09 mph. Compare with slip
claimed his speed at the 1/8 mile mark was 83.85 mph
Claim was about 1.45 % low
Regards - Ian
Speed v is dx/dt = (25.96884*1.54177) t^0.54177 ~ 40.0379784468 t^0.54177
but we do not yet know time t at the 1/8 mile mark.
Assuming distance x is based on feet, 1/8 mile = 5280/8 = 660 ft
so solve 25.96884t^1.54177 = 660 (on calculator) to find t
t = 8.1536 secs
v = 40.0379784468*(8.1536)^0.54177 ~124.7998 ft/sec
15 mph = (15*5280)/(3600) = 22 ft/sec
v = 124.7998 *(15/22) ~ 85.09 mph. Compare with slip
claimed his speed at the 1/8 mile mark was 83.85 mph
Claim was about 1.45 % low
Regards - Ian