An object has a position given by r = [2.0m + (3.0m/s)t] i + [3.0m - (2.0m/s^2)t

Question

An object has a position given by r = [2.0m + (3.0m/s)t] i + [3.0m - (2.0m/s^2)t^2] j Or (without units) r = [2.0 + (3.0) t] I + [3.0 - (2.0) t^2] j where all quantities are in SI units. What is the magnitude of the acceleration of the object at time 2.00 s? 1.00 m/s^2 0.00 m/s^2 0.522 m/s^2 4.00 m/s^2 2.00 m/s^2 A 6.0cm diameter crankshaft that is rotating at 2400rpm comes to a halt in 2.0s What is the tangential acceleration of a point on the surface? How many revolutions does the crankshaft make as it stops?

Explanation / Answer

2) r = (2 +3t) i +(3 -2t^2)j

velocity v = dr/dt

v = (3i) - (4t) j

acceleration a = dv/dt

a = -4 j

magnitude a = 4 m/s^2

correct option is (D)

3) d = 6 cm , r = 3 cm,

wo = 2400 rpm = 2400*2*3.14/60 = 251.2 rad/s

w = 0

t = 2 s

(a) alpha = (w-wo)/t

alpha = - 251.2/2 = -125.6 rad/s^2

at = r*alpha = 0.03*125.6

= 3.768 m/s^2

(b) from rotational kinematic equaitons

w^2 - wo^2 = 2*alpha*theta

0 - 251.2^2 = -2*125.6*theta

theta = 251.2 rad =251.2/(2*3.14)

theta = 40 rev

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