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Makovka662 [10]
3 years ago
15

Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take th

e speedometer readings every five seconds and record them in the following table. Time (s) 0 5 10 15 20 25 30 Velocity (mi/h) 16 21 23 26 33 30 28 In order to have the time and the velocity in consistent units, let's convert the velocity readings to feet per second (1 mi/h = 5280 3600 ft/s). (Roun
Physics
2 answers:
Reptile [31]3 years ago
8 0

Answer:

Distance traveled in 30 sec will be 1274.75 feet

Explanation:

We have given reading of velocity in mi per hour is

Velocity in mi per hour 16 21 23 26 33 30 28

Interval of time 0 5 10 15 20 25 30

And we know that 1 mi = 5280 feet and 1 hour = 3600 sec

So 1 mi/hr =\frac{5200}{3600}=1.44ft/sec

Now velocity in ft/sec 23.11  30.24  33.12  37.44  47.52  43.20  40.32

Time gap between each interval is 5 sec

So distance traveled will be =(23.11\times 5+30.24\times 5+33.12\times 5+37.44\times 5+47.52\times 5+43.2\times 5+40.32\times 5)=1274.75feet

Natalija [7]3 years ago
4 0

Let say for every 5 s of time interval the speed will remain constant

so it is given as

v(mi/h)   16    21    23    26    33    30     28

now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s

so here we will have

v(ft/s)      23.5    30.8     33.73     38.13     48.4     44     41.1

now for each interval of 5 s we will have to find the distance cover for above interval of time

d = v \times t

d = (23.5 + 30.8 + 33.73 + 38.13 + 48.4 + 44 + 41.1) \times 5

d = 1298.1 ft

so here it will cover 1298.1 ft distance in 30 s interval of time

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Arocket launches at an angle of 33.6 degrees from the horizontal at a
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