If a car covers 2/5th of the total distance with v1 speed and 3/5th distance with v2 than average speed is?
Call the total distance D and the total time T. (2/5)D is covered at speed v1. The time taken is t1 = distance/speed = (2/5)D/v1 (3/5)D is covered at speed v2. The time taken is t2 = distance/speed = (3/5)D/v2 The total time T is: T = t1 t2 . .= (2/5)D/v1 (3/5)D/v2 . .= (D/5)(2/v1 3/v2) . .= (D/5)(2v2 3v1) / (v1v2) Average speed = D/T = D / [ (D/5)(2v2 3v1) / (v1v2) ] = 5v1v2 / (2v2 3v1)
If a car covers 2/5th of the total distance with v1 speed and 3/5th distance with v2, then what is the average speed?
Suppose, the total distance to be covered by the car is xSo,time taken in covering 2/5 th of the total distance will be, t=(2x/5)/v1 (using, s=vt)Similarly, for the rest of the journey, time taken will be, t' =(3x/5)/v2Now, average speed = total distance covered/total time taken =x/(t+t')Putting the values we get, Vav =(5v1v2)/(2v2+3v1)
If a car covers2/5 of total distance with v1 speed and 3/5 distance with v2 Speed than the average speed is?
ist you need to know avg speed is Total distance/total timeIn given questions it is mentioned that 2 by 5 of distance is travelled with V1 speed and 3by 5 of distance is travelled with V2 speedYou can think it as follow total distance is 5 unit and one travel 2 unit with v1 speed and 3 unit with V2 speedtoatal distance 5 unittotal time= 2/v1 + 3/v1avg speed = 5/(2/v1 +3/v2)= 5v1 v2/(2v2+3v1)
A car covers the first half of a certain distance with a speed v1 and the second half with a speed v2.?
V*T = D. Since each velocity is over half the distance, then D1 = D2 V1 * T1 = V2 * T2 V(av) = D(total) / T(total) V(av) = (D1 + D2) / (T1 + T2) Let's assume D1 = 1. Then T = 1/V T1 = (1/V1) V(av) = (2) / (1/V1 + 1/V2) V(av) = 2 V1 V2 / (V1 + V2) Let's check that this seems correct with an example: V1 = 3, V2 = 10, D1=D2=1 T1 = 1/3, T2 = 1/10 D(total) = 1 + 1 = 2 T(total) = 1/3 + 1/10 = 10/30 + 3/30 => 13/30 V(av) = 2 / (13/30) => 60/13 V(av) = 2 V1 V2 / (V1 + V2) V(av) = 2 *3 * 10 / (3 +10) V(av) = 60/13
A particle covers each 1/3 of the total distance with speed v1, v2, v3. Find the average speed of the particle?
Total distance traveled = S Time taken for the 3 parts of the journey = t1, t2 and t3 respectively Speed during the 3 parts of the journey = v1, v2, v3, respectively Distance covered in part of the journey = S/3 => t1 = (S/3)/v1 = S/(3v1) t2 = (S/3)/v2 = S/(3v2) t3 = (S/3)/v3 = S/(3v3) Total time taken = t = t1 + t2 + t3 = (S/3)(1/v1 + 1/v2 + 1/v3) Average speed = V(av) = S/t = S/[(S/3)(1/v1 + 1/v2 + 1/v3)] = 1/[(1/3)(1/v1 + 1/v2 + 1/v3)] = 1/[(v2 v3 + v3 v1 + v1v2)/{3(v1 v2 v3)}] = (3 v1 v2 v3) / (v1 v2 + v2 v3 + v3 v1)
A car covers 1/3 part of total distance with speed of 20 km/hr & remaining distance with a speed of 44 km/hr. What is the average speed of the car?
1/3 of the distance is covered at 20 km/hr in t(1) time. 2/3 of the distance is covered at 44 km/hr in t(2) time. That means that the total time is 1/3 d ÷ 20 + 2/3 d ÷ 44. This gives us the equation t = d/60 + d/66. Simplified, this is t = 7d/220. As d = r•t, and d=(220/7)t, the average rate is 31 3/7 km/hr, approximately 31.42857 km/hr.
A car covers 2/5 the distance at a speed of 80 km/hr and 3/5 the distance at 60 km/hr. What is the average speed?
Let n be 1/5th of the distance covered. Then the first part is 2n long, and the second part is 3n long. The total time taken to cover these lengths is:2n/80 + 3n/60And the total distance is 5n. So, the average velocity is the total distance over the total time. Like this:V=5n/ 2n/80 + 3n/60=1200n/6n+12n=1200/18=66 2/3 km/h