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2 years ago

you drive in a straight line at 20.0m/s for 10.0 min, then 30.0m/s for another 10.0 min. Find the average speed.

Physics

1 answer:

Juliette [100K]2 years ago

6 0

Answer is 25 m/s.

Explanation;

Step 1 : Convert the units into same unit.

Here speeds are given in m/s while the time is in minutes. So, let's convert minutes into seconds.

1 min = 60 s

Then,

10 minutes = 10 x 60 s = 600 s

Step 2 : Find the total distance that you have travelled.

To find that you need to first find distances separately.

We know the formula, s = vt, where s is the distance, v is the speed or velocity and t is the time.

First you have travelled 10 minutes (600 s) at 20.0 m/s.

Then distance, s = 20.0 m/s x 600 s = 12 000 m

Secondly you have travelled again 10 minutes at 30.0 m/s

Then distance = 30.0 m/s x 600 s = 18 000 m

Then, the total distance = 12 000 m + 18 000 m = 30 000 m

Step 3 : Find the total time taken to travel.

Total time = 600 s + 600 s = 1200 s

Step 4 : Calculate the average speed.

Average speed = total distance / total time

= 30 000 m / 1200 s

= 25 m/s

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And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

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$2a_2(y_2-y_{02})=-v_{02}^2$

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