Answer:
v = 42.33 m/s
Explanation:
The time difference between two pulses that are emitted by Police is observed in the interval
[tex]\Delta t = 1.27 \times 10^{-7} s[/tex]
now the speed of the electromagnetic wave is given as
[tex]c = 3 \times 10^8 m/s[/tex]
now the distance traveled by them
[tex]d = c \times t[/tex]
[tex]d = (3 \times 10^8)(1.27 \times 10^{-7})[/tex]
[tex]d = 38.1 m[/tex]
so distance of speeding vehicle covered for round trip of wave is 38.1 m
now for one trip it is given as d = 19.05 m
now the speed of the vehicle is given as
[tex]v = \frac{d}{t}[/tex]
[tex]v = \frac{19.05}{0.450}[/tex]
[tex]v = 42.33 m/s[/tex]
The speed of the vehicle, is[tex]41.34m/s.[/tex]
The speed of an object serves as magnitude of the rate of change of its position with time.
Speed can be calculated as[tex]C= d/ t[/tex]
where [tex]C= speed= 3*10^8 m/s[/tex]
t= time
d= distance
The time interval between two pulses by the police= 1.27*10^-7 s.
[tex]Distance= ( 3*10^8 * 1.27*10^-7)= 38.1m[/tex]
To know distance traveled for a trip we will divide the distance traveled for a cycle which is [tex](38.1/2)= 19.05m[/tex][tex](38.1 /2) = 19.05m[/tex]
Then the speed of the vehicle can be calculated as C= d/ t
Where[tex]C= 19.05/0.450= 41.34m/s[/tex]
Therefore, speed of the vehicle, is [tex]41.34m/s.[/tex]
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