A car speeds up as it turns from traveling due south toheading due east. When ex

Question

A car speeds up as it turns from traveling due south toheading due east. When exactly halfway around the curve, the car'sacceleration is 2.70 m/s^2, 40.0 degrees north of east. What is the radial component of the acceleration at thatpoint? What is the tangential component of the acceleration at thatpoint? A car speeds up as it turns from traveling due south toheading due east. When exactly halfway around the curve, the car'sacceleration is 2.70 m/s^2, 40.0 degrees north of east. What is the radial component of the acceleration at thatpoint? What is the tangential component of the acceleration at thatpoint?

Explanation / Answer

The velocity of the car is v2 - u2 = 2aS u = 0,a = 2.70 m/s2 and S = r1 * where r1 = (r/2) and = radian = 3.14radian or v2 - 02 = 2 * 2.70 * r1 * = 2 * 2.70 * (r/2) * 3.14 = 8.478 * r or v2 = 8.478 * r The radial acceleration is ar = (v2/r1) = (8.478 *r/(r/2)) = 16.956 The tangential component of the acceleration at thatpoint at = r1 * ----------(1) we know that w2 - wo2 =2 or = (w2 -wo2/2) w = (v/r1),wo = 0 and = radian = 3.14 radian or = ((v/r1)2 -02/2) = (v2/2 *r12) = (v2/2 *(r/2)2) = (4v2/2 * r2)----------(2) From equations (1) and (2) we get at = (r/2) * (4v2/2 *r2) = (v2/ * r) = (v2/r) *(1/) or at = 8.478 * (1/3.14) = 2.70 m/s2 The tangential component of the acceleration at that pointis at = at * sin = 2.70 *sin(40.0o) = 1.73 m/s2 The tangential component of the acceleration at that pointis at = at * sin = 2.70 *sin(40.0o) = 1.73 m/s2

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