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The velocity of a car moving on a straight road increases linearly according to equation, v = a + bx, where a & b are positive constants. The acceleration in the course of such motion: (x is the displacement)
- a)Increases
- b)Decreases
- c)Stay constant
- d)Becomes zero
Correct answer is option 'A'. Can you explain this answer?
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Is the initial velocity and b is the acceleration.
To find the displacement of the car as a function of time, we need to integrate the velocity function with respect to time.
v = a bx
dx/dt = v
dx = v dt
Integrating both sides:
∫dx = ∫v dt
x = ∫(a bx) dt
x = a ∫(bx) dt
x = a (1/2 bxt^2 + C)
where C is the constant of integration.
To find C, we need to use the initial condition that x = 0 when t = 0.
x = a (1/2 bxt^2 + C)
0 = a (1/2 bx(0)^2 + C)
0 = aC
C = 0
So the displacement of the car as a function of time is:
x = 1/2 bxt^2 + a
Note that this is a quadratic function of time, which makes sense since the velocity is increasing linearly.
To find the displacement of the car as a function of time, we need to integrate the velocity function with respect to time.
v = a bx
dx/dt = v
dx = v dt
Integrating both sides:
∫dx = ∫v dt
x = ∫(a bx) dt
x = a ∫(bx) dt
x = a (1/2 bxt^2 + C)
where C is the constant of integration.
To find C, we need to use the initial condition that x = 0 when t = 0.
x = a (1/2 bxt^2 + C)
0 = a (1/2 bx(0)^2 + C)
0 = aC
C = 0
So the displacement of the car as a function of time is:
x = 1/2 bxt^2 + a
Note that this is a quadratic function of time, which makes sense since the velocity is increasing linearly.
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The velocity of a car moving on a straight road increases linearly according to equation, v = a + bx, where a & b are positive constants. The acceleration in the course of such motion: (x is the displacement)a)Increasesb)Decreasesc)Stay constantd)Becomes zeroCorrect answer is option 'A'. Can you explain this answer?
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