Physics CH 2 HW.pdf - (2-4 A rolling ball moves from x1 = 8.4cm t x2 = −4.2cm during the time from t1 = 3.0s to t2 = 6.1s What is its average velocity

# Physics CH 2 HW.pdf - (2-4 A rolling ball moves from x1 =...

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(2-4) A rolling ball moves from x1 = 8.4cm t x2 = 4.2cm during the time from t1 = 3.0s to t2 = 6.1s. What is its average velocity over this time interval? average v= displace/time= x2-x1/t2-t1= =-4.2cm-8.4cm/6.1s-3.0s= -12.6cm/3.1s = -4.06cm/s (2-11) A car traveling 95 km/h is 210 m behind a truck traveling 75 km/h. How long will it take the car to reach the truck? x=210m v=20km/h=5.555m/s t= x/v t=210m/5.555m/s= =37.804s (2-12) Calculate the average speed and average velocity of a complete round trip in which the outgoing 250km is covered at 95km/h, followed by a 1.0 hour lunch break, and then return 250km is covered at 55km/h. toutgoing=d/v= 250km/95km/h= 2.63h tlunch= 1h treturn=d/v=250km/55km/h= 4.54h t=toutgoing+tlunch+treturn= 2.63h+1h+4.54h= 8.17h average speed= (250km+250km)/8.17h= =61.15km/h vaverage=doutgoing+doutgoing+dreturn/toutgoing+tlunch+treturn= = (250km)+(0km)+(-250km)/8.17h= =0km/h (2-18) A sprinter accelerates from rest to 9.00m/s in 1.38s. What is her acceleration in (a) m/s2 u=0 a=v/t= 9.00m/s/1.38s= = 6.52m/s^2 (b) km/h2 6.52/1000 x (3600)^2= = 84499.2km/h^2 (2-19) A sports car moving at constant velocity travels 120m in 5.0s. If it then brakes and comes to a stop in 4.0s, what is the magnitude of its acceleration (assumed constant) in m/s2,

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