On a shopping trip, it took Audrey 1500.0 seconds to move from A to B, a distanc

Question

On a shopping trip, it took Audrey 1500.0 seconds to move from A to B, a distance of 50.0 m, she shopped for 26.0 seconds in B, then walked with a speed of 0.5 m/s from B to C (a distance of for 20.0 m), and with the same speed from C to D (a distance of 31.0 m).

a) What was in m/s the magnitude of the average velocity of audrey when she went from A to B?

b) How long, in seconds, did it take audrey to make it from A to D?

c) What is in m the magnitude of the overall displacement when audrey goes from A to D?

d) What is in m/s the magnitude of the average velocity for this shopping trip?

Explanation / Answer

Time taken from A to B=t1=15sec

S1=50m

time at B=26s

S2=20m

v2=0.5m/s

v3=o.5m/s

s3=31m

Requirements

A) total time = t=?

B) S= ?

C) Vav=?

Solution.

A Total time = t1+t2+t3+t4

= 1500sec+26sec+ 20/0.5ses+31/0.5sec

=1628sec

S= s1+s2+s3

S=50m+20m+31m

S=101m

V=(50/1500+0.5+0.5) /3

vav=o.344m/s

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