1. A car undergoes a displacement of 500m due east, followed by another displace

Question

1. A car undergoes a displacement of 500m due east, followed by another displacement due south of 1200m.
a) Calculate the total distance traversed by the car

A)St=s1+s2=500+1200=1700M
b) Calculate the total displacement of the car

B) D=(500m^2+1200m^2) sqr root= 1300

Tan-1(1200/500)=67.4 (angle)

2. If the car in problem 1 traveled at 36.0km/h during the first part of the trip and at 90.0km/h during the remainder,
a) Calculate the average speed for each part of the trip
b) Calculate the average velocity for each part of the trip
c) Calculate the time for each part of the trip
d) Calculate the average speed for the whole trip
e) Calculate the average velocity for the whole trip
f) Calculate the instantaneous speed of the car as it passes the 1km point of the trip
g) Calculate the average velocity of the car as it passes the 1km point of the trip

H) Calculate the average velocity required for the car in problem 2 to return to its starting point in the shortest possible distance in 98.0s.  

PLEASE answer question # 2 (a-h)

Explanation / Answer

a) s1= 36.0km/h= 10m/s and s2=90.0km/h= 25m/s

b) v1= (36.0km/h)i =(10m/s)i and s2= (90.0km/h)j = (25m/s)j

c) t1=d1/s1 = 500/10 = 50s and t2= 1200/25 = 48s

d) Save = distance /tTotal = (1700/98) = 17.35 m/s

e) |Vave | = displacement/tTotal = (1300/98) = 13.27 m/s j

Vave = Vavvex + Vavey = [(13.27 cos67.4)i + (13.27sin67.4)j] m/s = [(13.0)i + (3.12)j] m/s

f) After 1km car is moving with speed s2=90.0km/h= 25m/s

g) tinst = 1000/25 = 40s

Vave = displacement/tTotal = (1000/40) = 25 m/s

h) Vave = Vavvex + Vavey = [(13.27 cos67.4)i + (13.27sin67.4)j] m/s = [(13.0)i + (3.12)j] m/s

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