a particle travels 10 m in first 5 secs, 10 m in next 3 secs. assuming constant acceleration, what is the distance travelled in next 2 sec?
vivekkumarbhand:
Let assume initial velocity is u and constant acceleration a. using s=ut+.5 a t^2 For first 5 second 10=5 u +.5x a x 25..........(1) for next 3 seconds t=8 and s=20 20=8 u+.5xax64...........(2) Solving 1 and 2 a=1/3 and u=7/6 Now for next 2 seconds t=10 and s' S'=10x7/6 +.5x(1/3) 100 S'=170/6 Now distance travelled in 2 sec=(170/6)-20=50/6 meter
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Let assume initial velocity is u and constant acceleration a.
using s=ut+.5 a t^2
For first 5 second
10=5 u +.5x a x 25..........(1)
for next 3 seconds t=8 and s=20
20=8 u+.5xax64...........(2)
Solving 1 and 2
a=1/3 and u=7/6
Now for next 2 seconds t=10 and s'
S'=10x7/6 +.5x(1/3) 100
S'=170/6
Now distance travelled in 2 sec=(170/6)-20=50/6 meter
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