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dragster acceleration
For the dragster whose wheels roll without slipping, is there a relationship between the magnitude a (in m/s2)
Two identical dragsters, starting from rest, accelerate side-by-side along a straight track. The wheels on one of the cars roll without slipping, while the wheels on the other slip during part of the time.
For the dragster whose wheels roll without slipping, is there a relationship between the magnitude a (in m/s2) of its linear acceleration and the magnitude α (in rad/s2) of the angular acceleration of its wheels?
a) Yes. The relationship is a= alpha^2 / r, where r (in m) is the radius of a wheel.
b) Yes. The relationship is alpha= ra, where r (in m) is the radius of a wheel.
c) no
d) Yes. The relationship is a= r alpha, where r (in m) is the radius of a wheel.
d. Just as with velocity or displacement, every (rad/s^2, rad/s, rad) results in the rim motion of r times that quantity. By definition, the arc length of 1 radian is equal to the distance of the arc from the center, i.e., the radius.
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Drag Mustang with 1112 hp
An 893 kg (1974 lb) dragster, starting from rest...?
An 893 kg (1974 lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9 mph) in 0.59 s.
(a) Find the average acceleration of the dragster during this time interval.
___m/s2
(b) What is the magnitude of the average net force on the dragster during this time?
___N
(c) Assume that the driver has a mass of 68 kg. What horizontal force does the seat exert on the driver?
___N
(a)
a = dv/dt = (26.3 m/s) / (0.59 s) = 44.576 m/s²
(b)
F = ma = (893 kg)(44.576 m/s²) = 39806.37 N = 39.81 kN
(c)
F = ma = (68 kg)(44.576 m/s²) = 3031.17 N
Tags: acceleration · bedfordshire · cars · dragster acceleration · quarter-mile · racing · speed · top fuel dragster acceleration · top thrill dragster accelerationNo Comments
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