Physics, 30.09.2019 19:00 2023greenlanden

An unmarked police car traveling a constant 95 km/h is passed by a speeder. precisely 1.00 s after the speeder passes, the police officer steps on the accelerator. if the police car accelerates uniformly at 2.00 m/s2 and overtakes the speeder after accelerating for 7.00 s , what was the speeder's speed?

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Answer from: nama34

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So $90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec$

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with a=2m/sec^2 for 7 sec

So distance traveled by car in these 7 sec

$S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m$

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so $224=\left ( 1+7 \right )v$

v = 28 m/sec

Answer from: jadenp23

34.56m/s or $124.4kmhr^{-1$

Explanation:

Let the speed of the police car be v_p and that of the speeder be v_s

given; $v_p=95kmhr^{-1}=26.39m/s$ . This is the constant speed of the police car at the instant when the speeder overtook it.

Recall that generally $speed=\frac{distance}{time}...........(1)$

hence; distance = speed x time.

In 1s, the police car would have travelled a distance d = 26.39 x 1 =26.38m;

also in this time interval the speeder would have travelled a distance of

d_s=v_s x 1 = v_sm .

Therefore the differential distance between the speeder and the police car after 1s is given by;

D=v_s-26.39

At this instant, the police car begins to accelerate for 7s while the speeder maintained its constant speed for the same time. The police car would overtake the speeder at an instant when the distances travelled by both vehicles are equal, and this occurs after 7s as specified by the question.

Since the police accelerates for the 7s, we use the second equation of motion to get the distance covered as follows;

$s=ut+\frac{at^2}{2}$

Where;

$u= v_p=26.39m/s\\a=2m/s^2\\t=7s\\s=distance=?$

By substituting we obtain the following;

$s=(26.39*7)+\frac{2*7^2}{2}\\s=233.73m$

Since the speeder maintained constant speed, equation (1) will be used to determine its distance for the 7s as follows;

distance = speed x time = v_s*7=7v_s

total distance travelled by police car in 7s = 233.73 + D

Recall that D = v_s-26.39 ;

Also we said that overtaking occurs at the instant when the distances of both vehicles are equal, therefore;

$7v_s=233.73+D\\7v_s=233.73+v_s-26.39\\simplifying;\\6v_s=207.34\\$

$v_s=\frac{207.34}{6}=34.56m/s$

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

Answer from: Quest

it has to be b, about 97 percent.

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