If two cars are speeding at the same speed one right behind the other, which is more likely to get clocked and pulled over by the police? Why?
I can answer this because I was that very car. This is my only ever points on my license in all my driving career of over 2 million miles. Some ask, how do you know you've drove 2 million miles it seems a lot and unlikely.I was a same day courier and I very easily drove 2,000 miles a week. Over a year that's 100,000 miles and I've been driving over 20 years. But I haven't been driving that sort of distance since my Back Operatio, but it's still over 20 years.So there's my 2 million miles and like I said I can answer this question. I was driving through the heart of Wales on a single carriageway national speed limit, so 60 Mph. I was doing 81 Mph and there was a car behind me matching my speed. He had been behind me for a couple of miles when I glanced in my dirty mirror and saw a 3rd car quickly join us 2. Because my mirror was dirty I didn't make out it was a police car and the conditions were not ideal with slightly rainy weather. The road was a long wide downhill type with huge fields either side, so it wasn't like I didn't take the conditions into account, I did and I was very comfortable driving at that speed.Next thing I see is the blue lights go on and he gets in front of us both and we both stop. He walks up to the second car and within 30 seconds let's him go. I good sign I think…..He then proceeds to lecture me saying this and saying that and gives me a speeding ticket. So I ask, how come the other driver didn't get one and he tells me because I was the lead car, so it was my responsibility to adhere to the speed limit, he was just following???????I still don't quite know why he didn't get done and I did and I recall at the time ( over 20 yrs ago now ) I made a complaint, but never had anything back and just left it at that.I don't know if this is the norm, but I felt we both should of been done for speeding. The excuse the officer told me about me being the lead car to me was just bollox, but I don't know if this is true.
A car is going 60 MPH. How fast is that in ft/sec?
That is 88 ft per second The number is a constant I memorized when I used to do word problems in physics. It just pops into my head when I see a mph to fps question.
Suppose an object is moving at 66 ft/sec. How fast would you have to drive a car to keep pace with this object?
66ft/s = 45mph https://www.google.co.uk/webhp?source=se... http://www.calculateme.com/Speed/Feetper...
A car is travelling on a circular road of radius 500m. At some instant its speed is 30m/s and is increasing at the rate of 2m/s. What is its acceleration?
The acceleration has 2 components : tangential ([math]a_t[/math]= 2 m/s[math]^2[/math]) and centripetal ([math]a_r=\tfrac{v^2}{r}=\tfrac{30^2}{500}=1.8[/math]m/s[math]^2[/math]).Net acceleration = [math]\sqrt{2^2+1.8^2}[/math]m/s[math]^2[/math] = 2.69 m/s[math]^2[/math].
A car travels 200km in 2 hours and 24 minutes. What is the average speed that the car is traveling?
To solve this problem, we’ll utilize the “time-speed-distance” relationship which is expressed by the formula d = rt, i.e., distance is equal to speed multiplied by time.We’re given that d = 200 km and t = 2 hours and 24 minutes = (2 and 24/60) hours = 2⅖ hours = 2.4 hours. Now, substitute the given value for d and the given value for t into the time-speed-distance formula and then solve for speed r as follows:d = rt200 km = r(2.4 hours)r(2.4 hours) = 200 km (Equality is symmetric, i.e. if a = b, then b = a.)Let’s get rid of the decimal, i.e., 2.4, by multiplying both sides of the equation by 10:r(2.4 hours)(10) = (200 km)(10)r(24 hours) = 2000 kmNow, divide both sides of the equation by “24 hours” to isolate the unknown speed r by itself on the left side:[r(24 hours)]/24 hours = 2000 km/24 hoursr(24 hours/24 hours) = 2000 km/24 hoursr(1) = 2000 km/24 hoursOn the right side of the equation, divide both the numerator and the denominator by 24 as follows:r = (2000/24)km/(24/24) hourr = (2000/24) km/1 hourr = (2000/24) km/hr.r = 83⅓ km/hr. is the average speed that the car is traveling.
A car travels from A to B at the rate of 40 miles per hour, and then returns from B to A at the rate of 60 miles per hour. Is the average rate for the round trip more or less than 50 miles per hour? Why?
it 'll be 48 mph.let, the distance of A to B be 1mile.so,first the time=1/40hr.on return,it'll be 1/60hrso, in all for 2 mile the time=1/40+1/60hr.so, avg=2/(1/40+1/60)=48.