Question

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?

Answer 1

Answer:

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 2

Answer:

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;

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

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