Question

A car is moving with a velocity (3.0 m/s) x̂ + (1.0 m/s) ŷ and 3.0 seconds later its velocity is (6.0 m/s) x̂ - (3.0 m/s) ŷ. What is the direction of the average acceleration of the car?

Answer 1

Here we know that the initial velocity of the car is given by:

$V_i=(3.0\hat x+1.0\hat y)m/s$

And the final velocity of the car is given by:

$V_f=(6.0\hat x-3.0\hat y)m/s$

It took 3 seconds to attain the final velocity, so we have $t=3 s$

Therefore, the acceleration can be obtained by:

$V_f-V_i=a\times t$

$a=\frac{V_f-V_i}{t}$

Plugging the values of the initial, final velocity and the time, we get:

$a= \frac{(6\hat x-3\hat y)-(3 \hat x+1 \hat y)}{3} =\frac{6\hat x-3 \hat y -3 \hat x-1 \hat y}{3} =\frac{3 \hat x-4 \hat y}{3}$

So the acceleration of the car is given by:

$a=\frac{3 \hat x}{3} -\frac{4 \hat y}{3} =1 \hat x- \frac{4}{3} \hat y$

Now we need to find the direction of the average acceleration of the car:

$\theta=tan^{-1} \frac{y}{x}$

Here, x and y are the coefficients of the 'x' and 'y' components of the vector:

$\theta=tan^{-1}(\frac{-\frac{4}{3} }{1} )=tan^{-1}(\frac{-4}{3}) =-53.13^\circ$

Therefore, the direction of the average acceleration of the car is $-53.13^\circ$.