Respuesta :
Here we know that the initial velocity of the car is given by:
[tex]V_i=(3.0\hat x+1.0\hat y)m/s[/tex]
And the final velocity of the car is given by:
[tex]V_f=(6.0\hat x-3.0\hat y)m/s[/tex]
It took 3 seconds to attain the final velocity, so we have [tex]t=3 s[/tex]
Therefore, the acceleration can be obtained by:
[tex]V_f-V_i=a\times t[/tex]
[tex]a=\frac{V_f-V_i}{t}[/tex]
Plugging the values of the initial, final velocity and the time, we get:
[tex]a= \frac{(6\hat x-3\hat y)-(3 \hat x+1 \hat y)}{3} =\frac{6\hat x-3 \hat y -3 \hat x-1 \hat y}{3} =\frac{3 \hat x-4 \hat y}{3}[/tex]
So the acceleration of the car is given by:
[tex]a=\frac{3 \hat x}{3} -\frac{4 \hat y}{3} =1 \hat x- \frac{4}{3} \hat y[/tex]
Now we need to find the direction of the average acceleration of the car:
[tex]\theta=tan^{-1} \frac{y}{x}[/tex]
Here, x and y are the coefficients of the 'x' and 'y' components of the vector:
[tex]\theta=tan^{-1}(\frac{-\frac{4}{3} }{1} )=tan^{-1}(\frac{-4}{3}) =-53.13^\circ[/tex]
Therefore, the direction of the average acceleration of the car is [tex]-53.13^\circ[/tex].
The direction of the average acceleration of the car is -53.13 degrees.
We have to determine
What is the direction of the average acceleration of the car?
Acceleration
An object is said to be accelerated if there is a change in its velocity.
The acceleration of the car is given by;
[tex]\rm Acceleration=\dfrac{Change \ in \ velocity}{Time}\\\\ Acceleration=\dfrac{3x+1y-6x+3y}{3}\\\\ Acceleration=\dfrac{-3x+4y}{3}\\\\ Acceleration=\dfrac{-3x}{3}+\dfrac{4y}{3}\\\\ Acceleration=-x+\dfrac{4y}{3}\\\\[/tex]
The direction of the average acceleration of the car is;
[tex]\rm \theta=tan{-1}\dfrac{y}{x}\\\\\theta=tan{-1}\dfrac{4/3}{1}\\\\ \theta=tan^{-1}\left ({\dfrac{-4}{3} \right )\\\\ \theta=-53.13[/tex]
Hence, the direction of the average acceleration of the car is -53.13 degrees.
To know more about acceleration click the link given below.
https://brainly.com/question/3548817
