Question

12-B12 This car is driving up the hill shown. When he passes the origin he is travelling 35 m/s and thereafter his x velocity is defined as Vx=35-1.2t. At t=15 seconds, determine The position (x,y) and the x,y components of both velocity and acceleration. • What does his speedometer read at that point in km/hr? y = 0.0003.r

Answer 1

- Given dy da (chain rule) velocity at Rigin voz 25 m/s V x = 35-112t, 420.00037, vyzdy Jun 25-1-2t S » * f35-1-2>> Sdn 525-1.27 35t-1-2t+C 1 2 35t-1.2t4 at no t=0, so czo at t= 15 sec & 235x15- 1.2 (15) 2390m, y 0.0003 * 2 0-0003 (390) 245.63m ay a dry dux der ay 2 (0.0006)x(-12) Tay 27.2800 m15 Net velocity vo svettvy (n.) 2 (390, 45.63] - (VWt 18 2 35-1-2 (15) 2 17 m/s (vy) 2 x -00064){25=128) 16.000 /25 - 12) V2 √ (17)²+(3-978) 2 V 217.459 m/s km/h De=is 26.04(8st-124"><C25-12) (0.006 (25+15-12) (15.1.2318 | v2 17.459x360 Iva 62.853 km/hr Tuy 23. 978 m/s] Speedometee reading & 62.853 km/hr acceleration dva - are so 8-1.2 m/s² at tois, ar 2-1.2 m/s2