A rabbit runs across a parking lot .The path is such that components of rabbit's position with respect to an origin of co ordinates as function of time are x= t^2 +10 and y=-9t +5 .here X ,y are in meter and t in second . the magnitude of rabbit displacement during first 5 seconds
Answers
Magnitude of rabbit's displacement during first 5 seconds is 51.5 m
Explanation:
Given:
x = t² + 10
y = -9t + 5
Distance in meters and t is seconds.
When t = 0, x and y will be:
x = 0² + 10 = 10
y = -9(0) + 5 = 5
(10 , 5)
When t = 5 seconds, x and y will be:
x = 5² + 10 =35
y = -9(5) + 5 = -40
(35 , -40)
Displacement = distance between the x and y coordinates when t = 0 and t - 5s.
Displacement = √(35 - 10)² + (-40 - 5)²
= √(25)² + (-45)²
= 5(10.3)
= 51.5 m
Hence, the magnitude of rabbit's displacement during first 5 seconds is 51.5 m
Given : A rabbit runs across a parking lot. The path is such that components of rabbit's position with respect to an origin of co-ordinates as function of time are x = t^2 + 10 and y = -9t + 5.
Here x, y are in meter and t in second.
To Find : The magnitude of rabbit's displacement during first 5 seconds is
a) 70.5 m
b) 51.5 m
c) 62.5 m
d) 65.5 m
Solution:
x = t² + 10
y = -9t + 5
at t = 0
x = 0² + 10 = 10
y = -9(0) + 5 = 5
(10 , 5)
at t = 5
x = 5² + 10 =35
y = -9(5) + 5 = -40
(35 , -40)
Displacement = √(35 - 10)² + (-40 - 5)²
= √(25)² + (-45)²
= 5√(25 + 81)
= 5√106
= 5 x 10.3
= 51.5 m
The magnitude of rabbit's displacement during first 5 seconds is 51.5 m
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