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andrew-mc [135]
3 years ago
11

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. Wha

t is the direction of the average acceleration of the car?
Mathematics
1 answer:
Mars2501 [29]3 years ago
8 0

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.

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