A car travels a distance of 2000 M if the first half distance is covered at 40 km per hour and the second half with the speed v and if the average velocity is 48 km per hour then the value of V
Answers
Answer:V(avg) = [d1 + d2] /[t1 +t2]
48 = [d/2 + d/2] / [(d/2)/40 +(d/2)/V]
48 = 2 / [1/40 + 1/V]
48 = 80V/40+V
48V + 48(40) = 80V
1920= 32V
V = 60km/h
Explanation:
Because the average speed for the whole trip is higher than the speed for the first half of the trip, it’s obvious that the speed during the second half of the trip is higher than the first, and the time taken for the second half is shorter. Because the durations of the 2 phases of the trip are different, we CANNOT assume that we can simply average the speed out.
Let’s go back to basics. Since Speed = Distance Travelled divided by Time Taken,
then time = dist/speed.
For the overall journey, the average speed is 48km/h over a distance of 2000m. But 48km/h = 48000metres/3600seconds = 40/3 m/s. Note that I’ve assumed that there was no change in direction, so the speed is effectively the scalar component of the car’s velocity.
So for the overall journey, T = dist/speed = 2000m / (40/3 m/s) = 2000 * 3/40 s = 150 seconds.
During the first part of the journey, we are told that the car travelled 1000 metres at a speed of 40km/h. By a similar calculation, the time taken for this phase of the journey was 90 seconds.
We now know that the time taken for the second part of the journey was 60 seconds (150 - 90 seconds) and the distance travelled during this part was 1000 metres.
Therefore the speed during the second half of the journey was (dist/time) 1000m/60s which converts nicely to 60km/h.
v = 60km/h.