Answer:
V(car) = V(truck) at t = Dt/2
acceleration = v(car) = D/t^2
Explanation:
acceleration = v(car) = D/t^2
Since the average velocities must be the same, the car's final velocity must be twice the trunk velocity assuming the car start with zero velocity, since acceleration remain the same throughout the journey velocities at half-time point must be equal.
A body is accelerating when its velocity increases with time
Part A:
[tex]The \ acceleration \ of \ the \ car, \ a = \mathbf{\dfrac{2\times \left(v(t) \right)^2}{D}}[/tex]
Part B:
[tex]\mathbf{The \ speed \ of \ the \ car \ equals \ the \ speed \ of \ the \ truck \ at \ time}, t = \mathbf{\dfrac{ t_D }{2}}[/tex]
The reason the above expressions are correct is as follows:
Question: The introductory part of the question that appear missing is as follows;
A car is accelerating at a constant rate from rest at time t = 0, and it is about to be overtaken by a truck moving at constant velocity, v(t) such that the front of the car and the truck are are aligned or level at the same position at point x = 0. The car maintains its acceleration and after some distance is about to overtake the truck such that at point x = D, the front of the truck and the car are level once more
Part A: Known parameters
From the question, we have;
The velocity of the car at time t = 0 is v₁ = 0
The constant velocity of the truck = v(t)
Let the velocity of the car at time t = D = v₂
The car and the truck travel the same distance in the same time, therefore;
Average velocity of the car = Average velocity of the truck
[tex]Average \ velocity = \mathbf{\dfrac{Total \ distance}{Time}}[/tex]
Given that the distance = D - 0 = D, we have;
[tex]Average \ velocity\ \mathbf { = \dfrac{D}{t}}[/tex]
Where the velocity is constant, as in the truck, we have;
Average velocity = The constant velocity = v(t)
Therefore;
[tex]\mathbf{\dfrac{D}{t}} = v(t)[/tex]
[tex]t = \mathbf{\dfrac{D}{v(t)}}[/tex]
When the acceleration is constant, such as for the car, we have;
[tex]Average \ velocity = \mathbf{ \dfrac{Initial \ velocity + Final \ velocity}{2}}[/tex]
The above equation for the car, gives;
[tex]Average \ velocity = \dfrac{v_2 +v_1}{2} = \dfrac{v_2 +0}{2} = \dfrac{v_2 }{2}[/tex]
Therefore;
[tex]\dfrac{D}{t} = \dfrac{v_2 }{2}[/tex]
[tex]v_2 = \dfrac{2 \times D}{t}[/tex]
The acceleration of the car, a, is given using the constant acceleration formula as follows;
[tex]a = \dfrac{v_2 - v_1}{t} = \dfrac{v_2 - 0}{t} = \mathbf{\dfrac{v_2}{t}}[/tex]
Therefore;
[tex]a = \dfrac{v_2}{t} = \dfrac{\dfrac{2 \times D}{t} }{t} = \dfrac{2 \times D}{t^2} = \dfrac{2 \times D}{\left(\dfrac{D}{v(t)} \right)^2} = \mathbf{\dfrac{2\times \left(v(t) \right)^2}{D}}[/tex]
The acceleration of the car, a, in terms of v(t) and D is therefore;
[tex]a = \mathbf{\dfrac{2\times \left(v(t) \right)^2}{D}}[/tex]
Part B:
Let t represent the time at which the speed of the car is equal to the speed of the truck, we have;
a·t = v(t)
D = tD × v(t),
Therefore;
[tex]a \cdot t = \dfrac{2\times \left(v(t) \right)^2}{t_D \times v(t)} \times t = v(t)[/tex]
[tex]t = \dfrac{v(t) \times t_D \times v(t)}{2\times \left(v(t) \right)^2} = \dfrac{ t_D \times \left(v(t) \right)^2}{2\times \left(v(t) \right)^2} = \mathbf{\dfrac{ t_D }{2}}[/tex]
Therefore;
[tex]\mathbf{The \ time \ the \ speed \ of \ the \ car \ is \ equal \ to \ the \ speed \ of \ the \ truck}, \ t = \mathbf{\dfrac{ t_D }{2}}[/tex]
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