8.) Treat the graph below as a displacement vs time graph. a.) what is the avera

Question

8.) Treat the graph below as a displacement vs time graph. a.) what is the average velocity for each of the segments, A,B,and C? (-20m/s,10m/s,40m/s) b.) What is the instantaneous velocity at t-2s? (10m/s) c) What is the position at a time of t-o5s7(30m) +50.0 +40.0 +30.0 +20.0 +10.0 0 0.5 1.0 1.5 2.0 2.5 3.0 Time i S 10.)Below is a velocity vs time graph a.) What is the acceleration between 15 and 25? (0.5 b.) What is the total displacement between 0 and 15s? (125m) 15 10 (T 5 10 15 20 25 Time, t (s)

Explanation / Answer

From the given position vs time graph

velocity is by definition the rate of chage of velocity

from the graph the slope of the curve is the velocity

a) average velocity is v_avg = Dx/Dt

for the segment A , v_avg = (10-40)/(1.5-0) m/s = -20 m/s

for the segment B ,v_avg = (20-10)/(2.5-1.5) m/s = 10 m/s

for the segment c ,v_avg = (40-20)/(3.0-2.5) m/s = 40 m/s

b)instantaneous veloiscity at t = 2 s is  

the slope of the curve at t = 2 s

that is from t= 1.5 s tot= 2.5 s the slope is same which is equal to the velocity = (20-10)/(2.5-1.5) m/s = 10 m/s

c) from the given graph the position at t = 0.5s is x = 30 m  

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10

velocity vs time graph

the slope is the acceleration and the area under the curve gives the displacement

a) acceleration between 15 s and 25 s is  

a_avg = dV/dt = (5-10)/(25-15) m/s2 = -0.5 m/s2

b) area under the curve from 0s to 15 s

area = (0.5*5*10)+(10*10) = 125 m

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