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1.8 Problems
1.The position of a particle moving along the straight line is given in
2 3
meters by equation S  At  Bt  Ct , where t is in seconds,

A  4 m , B  3 m , C  2 m . Consider the time interval t  s 2
s s 2 s 3 1
tot  s 3 and calculate

1. average speed v
2. instantaneous speed at t  s 2 and t  s 3
1
2
3 .average acceleration a
4. instantaneous acceleration at t  s 2 and t  s 3
1
2

Solution
1.The ratio of the distance to the time, in which this distance was passed,

is average speed.
 S
 v 
t 
where S  S  S and t  t  t
2
2
1
1
Therefore
2 3 2 2
 S  (At  Bt  Ct 1 )  (At  Bt  Ct )
2
2
2
1
1
Substitute numerical value and obtain
2
2
 S  4 (  3  3 3  2 3  3 )  4 (  2  3 2  2 2  3 )  57 m
Then
m
v  57
s
2.Instantaneous speed v is the first derivative of distance with respect to

dS
time v 
dt
Therefore
2
v  A  2Bt  3Ct
According to the situation problem of instantaneous speed at time
t  s 2 is equal
1
m m m 2 m
v  4  2 3  2  s 3 2 2 (  s)  40
1 2 3
s s s s
Ditto to the time t  s 3
2
m m m 2 m
v  4  2 3  3  s 3 2 3 (  s)  76
2
s s 2 s 3 s

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