Please help me solve the following problem, thank you. A car is travelling at 76
ID: 1842954 • Letter: P
Question
Please help me solve the following problem, thank you.
A car is travelling at 76 km/h when it enters the semicircular curve at Point A. The car increases the speed at a uniform rate, exiting the curve at Point C with a speed of 108 km/h. The radius of the semicircular curve is R = 300m. Determine the velocity and acceleration of the car at Point B (normal/tangential coordinates). Determine the velocity and acceleration of the car at Point B (Cartesian coordinates) How long it takes for the car to pass from Point B to Point C?Explanation / Answer
>> Velocity at A, Va = 76 km/hr = 21.11 m/s
>> Velocity at C, Vc = 108 km/hr = 30 m/s
>> Displacement between B & C = 2*R = 600 m
>> Using Newton's 3rd Equation of Motion, v2 - u2 = 2ats
at = Tangential Acceleration
=> 30*30 - 21.11*21.11 = 2*at*600
Solving, at = 0.379 m/s2 ....Tangential Acceleration....
>> Now velocity at B = Vb
and, displacement between A & B = 20.5*R = 424.264 m
Using, 3rd Equation : v2 - u2 = 2ats
=> v2 - 21.11*21.11 = 2*0.379*424.264
=> Vb = 27.70 m/s = 99.72 km/hr
So, Centrepetal or Normal Acceleration at B, acb = Vb2/R = 27.70*27.70/300 = 2.56 m/s2 ...in Normal direction acting towards center ...
>> and,
at = 0.379 m/s2 ....Tangential Acceleration at B acting tangent to the curve in vertical downward direction
>> In cartesian Coordinate,
A = - 2.56 i - 0.379 j { m/s2 } .......ANSWER...
>> Part (c)
Applying Newton's 1st Equation of Motion,
V = U + a*t
=> 30 = 27.70 + 0.379*t
=> t = 6.10 sec ....REQUIRED TIME.....