The acceleration of a dragster is assumed to decrease as the speed increases. If

Question

The acceleration of a dragster is assumed to decrease as the speed increases. If a linear relationship is assumed a=ac - bv where the acceleration a is in ft / s2 and the velocity v is in ft / s. Using fundamental equations a = dv / dt and v= ds / dt Use the acceleration to determine the velocity as a function of time. Use the velocity to determine the distance traveled as a function of time For ac=350ft / s2 and b=0.5s-1, determine: the initial acceleration the elapsed time for the quarter mile the speed at the end of the quarter mile the acceleration at the end of the quarter mile Is this a record beating run? (3.58 sec and 386 mph)

Explanation / Answer

dv.dt = ac - bv

so, dv / (ac - bv) = dt

so, -1/b * ln (ac - bv) = t

so, a - bv = e^ (-bt)

so, V = (a - e^(-bt))/b

b) now. dx/dt = (a - e^(-bt))/b

so,dx = (a - e^(-bt))/b .dt

so, X = (a.t +b.e^(bt))/b

c) a = 350 - b x 0 (initially v = 0)

d) 1/4 mile = 1320 feet = (350 + 0.5 x e^(0.5 x t))/0.5

so, time = t = 12.86 s

e) V = (350 - e^ (0.5 x 12.86))/0.5 = - 540 m/s

f) a = 350 + 0.5 x 540 = - 620 ft/s^2

g) no. as time taken > 3.28 s

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