The velocity (v) of a particle that moves in the
positive x-direction varies with its position (x) as
shown in figure. The acceleration of particle at
x = 5 m is nearly
vms)
*(m)
(1) 10 ms2
(2) -9.4 ms?
(3) 75 ms
(4) -1.25 ms-2
Answers
Answer:
-9.4 m/s²
Explanation:
We know,
=> velocity (v) = dx/dt
=> acceleration (a) = dv/dt
let us derive another equation from this,
=> a = dv/dt
Multiplying both sides by by dx
=> a.dx = dv. dx/dt
=> a.dx = v.dv
Now integrating both sides we get,
=> a.(x2 - x1) = {(v2)² - (v1)²}/2
=> 2.a.(x2 - x1) = (v2)² - (v1)²
NOW,
To know,
=> if slope is a straight line at a certain angle , then acceleration at any point on that is same.
let us take readings from graph,
since 5 m lies between 3 and 7 m and also the slope is a straight line , so acceleration at 5 m is same to any point on that slope
=> x1 = 3 m
=> x2 = 7 m
=> v1 = 10 m/s
=> v2 = 5 m/s
By formula
=> 2.a.(7 - 3) = 5² - 10²
=> 2.a.4 = 25 - 100
=> 8a = -75
=> a = - 9.375 ≈ -9.4 m/s²
Hence the required acceleration is - 9.4 m/s²