Calculating Average Acceleration for Changing Velocity and Direction

[ognik]

I would just appreciate someone checking this please, I'm not sure of my answer ...

Q: Car moves North with constant speed 50 mph for 5 mins. It then turns 45 degree east and continues at 55 mph for 1 min. Find ave. acceleration.

For x, y components, Cos45 = Sin45 = 0.7, so for the 2nd part $v_x = v_y = 55 \times 0.7 = 38.9 $ mph for 1 min

So, $ a_x (ave) = \frac{\Delta v_x}{\Delta t} = \frac{38.9 - 0}{5 + 1}(60) = 389 $ and $ a_y (ave) = \frac{\Delta v_y}{\Delta t} = \frac{38.9 - 55}{5 + 1}(60) = -161 $ ?

 

[zzephod]

I would just appreciate someone checking this please, I'm not sure of my answer ...

Q: Car moves North with constant speed 50 mph for 5 mins. It then turns 45 degree east and continues at 55 mph for 1 min. Find ave. acceleration.

For x, y components, Cos45 = Sin45 = 0.7, so for the 2nd part $v_x = v_y = 55 \times 0.7 = 38.9 $ mph for 1 min

So, $ a_x (ave) = \frac{\Delta v_x}{\Delta t} = \frac{38.9 - 0}{5 + 1}(60) = 389 $ and $ a_y (ave) = \frac{\Delta v_y}{\Delta t} = \frac{38.9 - 55}{5 + 1}(60) = -161 $ ?

What units do you think your average acceleration has?

.

 

[ognik]

Hi - was not really worrying about getting the units strictly correct, just wanted to make sure I understood how to apply the average in this situation (where to me the average acceleration seems a bit meaningless)... but the units should be mph ph or $\frac{miles}{hour^2}$

 

[zzephod]

Average acceleration is change in velocity divided by the time. Here the initial velocity is $v_0=(0,50)$ and the final velocity is $v_1=(38.89,38.89)$ in units of mph, and the time interval is $6$ minutes or $1/10$ hours.

So the average acceleration is:
$$
\overline{a}=\frac{(38.89,38.89)-(0,50)}{1/10}=10(38.89,-11.11)=(388.9,111.1) \mbox{ mph/h}
$$

 

[ognik]

So my method seems ok thanks (I noticed I used 55 instead of 50 for the y component).