A particle moving in a straight line covers half the distance with speed 8 m/s. The other half of the distance is covered in two equal time intervals with speed 10 m/s and 14 m/s. respectively. The average speed of the particle during this motion is
Answers
10 m/s
First we have to calculate the average speed of the 2nd half distance.
Average speed in 2nd half distance = (10+14)/2 m/s
= 12 m/s
Now we have to calculate the average speed of 1st and 2nd half distance.
Speed in 1st half = 8 m/s
Speed in 2nd half = 12 m/s
Average speed = (8+12)/2 m/s
= 10 m/s
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Given data : A particle moving in a straight line covers half the distance with speed 8 m/s. The other half of the distance is covered in two equal time intervals with speed 10 m/s and 14 m/s respectively.
To find : The average speed of the particle during this motion ?
Solution : Let, total distance covered by partical be x.
Now, a/c to given data;
Speed of the particle = 8 m/s
Distance covered by the partical = x/2
Time taken by partical, T[1] = ?
By formula of speed;
➜ speed = distance/time
➜ 8 = (x/2)/T[1]
➜ T[1] = x/(2 * 8)
➜ T[1] = x/16 ----{1}
similarly {The other half of the distance is covered in two equal time intervals with speed 10 m/s and 14 m/s respectively}
Hence,
➜ x/2 = T[2] * 10 + T[3] * 14
Where, T[2] and T[3] are time intervals.
A/C to given data; T[2] = T[3]
Hence,
➜ x/2 = T[2] * 10 + T[2] * 14
➜ x/2 = 24 * T[2]
➜ T[2] = x/(2 * 24)
➜ T[2] = x/48
Hence, T[2] = x/48 and T[3] = x/48
Now,
➜ Total time taken = T[1] + T[2] + T[3]
➜ Total time taken = x/16 + x/48 + x/48
{Here, (x * 3)/(16 * 3) = 3x/48}
➜ Total time taken = 3x/48 + x/48 + x/48
➜ Total time taken = (3x + x + x)/48
➜ Total time taken = 5x/48
Now,
➜ Average speed = Total distance/total time taken
➜ Average speed = x/(5x/48)
➜ Average speed = 48x/5x
➜ Average speed = 9.6 m/s
Answer : The average speed of the particle during this motion is is 9.6 m/s.