Example 8.5 - Chapter 8 Class 9 - Motion
Last updated at April 16, 2024 by Teachoo
Examples from NCERT Book
Last updated at April 16, 2024 by Teachoo
Example 8.5 A train starting from rest attains a velocity of 72 km hβ1 in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity. Train starts from rest, Initial velocity = u = 0 km /h Final velocity = v = 72 km /h Time taken = 5 minute = 5 Γ 60 = 300 seconds Converting initial and final velocity in m/s Conversion from km/h to m/s 1 km = 1000 m 1 hour = 60 minutes = 3600 s β΄ (1 ππ)/β = (1000 π)/(3600 π ) β΄ 1 km/h = 5/18 m/s Initial velocity = u = 0 km/h = 0 Γ 5/18 m/s = 0 m/s Final velocity = v = 72 km/h = 72 Γ 5/18 m/s = 4 Γ 5 m/s = 20 m/s (i) Finding Acceleration Acceleration = (πΉππππ π£ππππππ‘π¦ β πΌπππ‘πππ π£ππππππ‘π¦)/(ππππ π‘ππππ) = (π£ β π’)/π‘ = (20 β 0)/300 = 1/15 m/s2 (ii) Finding Distance Travelled We know u, v, a and t Finding distance (s) by using 3rd equation of motion v2 β u2 = 2 as (20)2 β 0 = 2 Γ 1/15 Γ s 400 = 2/15 s 400 Γ 15/2 = s 200 Γ 15 = s 3000 = s s = 3000 m s = 3 km