STARTING @ 50.0m AWAY FROM A TELEPHONE POLE, A CAR ACCELERATED FROM REST @ A CONSTANT RATE OF 4.0m/s2?

1. HOW FAST WAS THE GOING AFTER 10s?


2. HOW FAR WAS IT FROM THE TELEPHONE POLL AFTER 10s?


3. WHAT WAS ITS AVERAGE VELOCITY FOR THE INTERVAL 0 TO 10s?|||1) v= u +at


v = 0+4*10


v = 40m/s





2) s = ut +½*a*t²


s = 0 + ½ *4*10²


s = 200m


He was 150m past the telephone pole or he could be 250 away from the pole depending on what is meant by "away from a telephone pole" Was the car 50m behind the pole or was the pole 50m behind the car





3) average velocity = displacement / time = 200m/10 = 20m/s|||by using integral calculus,





a = dv/dt = 4





dv = 4dt





∫dv = 4∫dt





v = 4t + v₀





wherein at t = 0, v₀ = 0 (since starting from rest)





v = 4t =%26gt; eq'n 1





v = ds/dt = 4t





ds = 4tdt





∫ds = ∫4tdt





s = 2t² + s₀





wherein at t = 0, s₀ = 50





s = 2t² + 50 =%26gt; eq'n 2





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1. speed was it going after 10 secs





v = 4(10)





v = 40 m/s





2. how far from the pole after 10 secs





s = 2t² + 50





s = 2(10)² + 50





s = 250 meters from the pole





3. average velocity from t = 0 up to t = 10





at t = 0, s = 0 and at t = 10, s = 200





average velocity = 200/10 = 20 m/s|||1. v=u+at


v=0+4x10


v=40m/s





2. s=ut+1/2at^2


s=0+1/2x4x10^2


s=200m


The car was 200-50=150m from the pole





3. Average Velocity=Total Distance/Total Time


AV=200/10=20m/s