
A particle has an initial velocity $ 4\hat i + 4\hat jm{s^{ - 1}} $ and an acceleration $ - 0.4\hat im{s^{ - 2}} $ , at what time will its speed be $ 5m{s^{ - 1}} $ ?
A. $ 2.5\;s $
B. $ 17.5\;s $
C. $ s\; $
D. $ 8.5\;s $
Answer
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Hint: The initial velocity of a particle is given as $ 4\hat i + 4\hat jm/s $ . There is an acceleration of $ - 0.4\hat im/{s^2} $ . We have to find the time at which the particle attains a speed of $ 5m/{s^2} $ . The acceleration of a particle is the change in the velocity of the particle. The negative acceleration is called retardation.
Complete step by step answer:
Here the particle is moving in the X-Y plane, the velocity of the particle in the X and Y direction is given. But we can see that the acceleration is only in the X-direction. Therefore, the velocity in the Y direction will not change. Only the velocity in the X-direction changes.
The given speed is $ 5m/s $ .
Then we can write,
$ {5^2} = V_x^2 + V_y^2 $
Let the velocity in the X-direction be $ {V_x} $
The velocity in the Y-direction is $ 4m/s $
Substituting,
$ {5^2} = V_x^2 + {\left( 4 \right)^2} $
From this, we can write,
$ {5^2} - {4^2} = V_x^2 $
That is
$ V_x^2 = 25 - 16 = 9 $
Taking the square root,
$ {V_x} = \pm 3m/s $
The velocity can be written as,
$ {V_x} = {u_x} + {a_x}t $
Where $ {u_x} $ is the initial velocity in the X-direction, $ {a_x} $ is the acceleration in the X-direction, and $ t $ stands for the time.
From this equation, we can write the time as,
$ t = \dfrac{{{V_x} - {u_x}}}{{{a_x}}} $
Let us first take the initial velocity to be $ + 3m/s $
$ \Rightarrow t = \dfrac{{3 - 4}}{{ - 0.4}} = \dfrac{{ - 1}}{{ - 0.4}} = 2.5m/s $
Now if the initial velocity is taken to be $ - 3m/s $
$ \Rightarrow t = \dfrac{{ - 3 - 4}}{{ - 0.4}} = \dfrac{{ - 7}}{{ - 0.4}} = 17.5m/s $
The time at which the speed is $ 5m/s $ can either be $ 2.5m/s $ or $ 17.5m/s $ depending on the velocity in X-direction.
Therefore, the answer is both option (A) and option (B).
Note: The velocity of an object at any instant is called the instantaneous velocity. It is the average velocity when the time interval tends to zero. The velocity is the time rate of displacement or the displacement in one second. Velocity can be negative, positive, or zero.
Complete step by step answer:
Here the particle is moving in the X-Y plane, the velocity of the particle in the X and Y direction is given. But we can see that the acceleration is only in the X-direction. Therefore, the velocity in the Y direction will not change. Only the velocity in the X-direction changes.
The given speed is $ 5m/s $ .
Then we can write,
$ {5^2} = V_x^2 + V_y^2 $
Let the velocity in the X-direction be $ {V_x} $
The velocity in the Y-direction is $ 4m/s $
Substituting,
$ {5^2} = V_x^2 + {\left( 4 \right)^2} $
From this, we can write,
$ {5^2} - {4^2} = V_x^2 $
That is
$ V_x^2 = 25 - 16 = 9 $
Taking the square root,
$ {V_x} = \pm 3m/s $
The velocity can be written as,
$ {V_x} = {u_x} + {a_x}t $
Where $ {u_x} $ is the initial velocity in the X-direction, $ {a_x} $ is the acceleration in the X-direction, and $ t $ stands for the time.
From this equation, we can write the time as,
$ t = \dfrac{{{V_x} - {u_x}}}{{{a_x}}} $
Let us first take the initial velocity to be $ + 3m/s $
$ \Rightarrow t = \dfrac{{3 - 4}}{{ - 0.4}} = \dfrac{{ - 1}}{{ - 0.4}} = 2.5m/s $
Now if the initial velocity is taken to be $ - 3m/s $
$ \Rightarrow t = \dfrac{{ - 3 - 4}}{{ - 0.4}} = \dfrac{{ - 7}}{{ - 0.4}} = 17.5m/s $
The time at which the speed is $ 5m/s $ can either be $ 2.5m/s $ or $ 17.5m/s $ depending on the velocity in X-direction.
Therefore, the answer is both option (A) and option (B).
Note: The velocity of an object at any instant is called the instantaneous velocity. It is the average velocity when the time interval tends to zero. The velocity is the time rate of displacement or the displacement in one second. Velocity can be negative, positive, or zero.
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