Consider the highway exit ramp with a radius rho = 0.25 mi with a car traveling

Question

Consider the highway exit ramp with a radius rho = 0.25 mi with a car traveling at 40 mph. Calculate the acceleration vector expressed in ft/s^2. A robot arm moves so that P travels in a circle about point B, which is not moving. Knowing that the speed of P increases from 15 mm/s to 40 mm/s in 2 sec. determine: The velocity vector when t = 1.5 sec. The acceleration vector when t = 1.5 sec. The magnitude of acceleration when t = 1.5 sec. For the child walking on the merry-go-round shown below, with a constant speed u = 2 ft/s and constant angular velocity of the merry-go-round of omega = 25 rpm. Assuming the child starts at the center at t = 0 sec. and walks in a straight line towards the outside: Calculate the velocity components at t = 3.5 sec. Calculate the acceleration components at t = 3.5 sec making sure to label each

Explanation / Answer

Radius = 0.25 miles = 1609×0.25 metre = 402 m

Velocity = 40 mph = 17.88 m/sec

So, acceleration = v2/r = 17.88×17.88/402 = 0.795 metre per second square = 2.61 ft/ sec2

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