Question

A body is moving with uniform accelaration goes 65 meter 5th second and 105 meter in 9th second. How far will it go in 20th seco d

Answer 1

at 4.5 seconds it is going 65 m/s

at 8.5 seconds it is going 105 m/s

a = change in v / change in time

a = 40/4 = 10 m/s^2
so
19.5 - 8.5 = 11 seconds

v at 19.5 = v at 8.5 + 10*11
= 8.5 + 110
= 118.5 meters/second at 19.5
that is average speed from t = 19 to t = 20
so
118.5 meters

Answer 2

using Calculus:

v = at + c
s = (1/2)at^2 + ct + k

when t = 4
s = 8a + 4c + k
when t = 5
s = (25/2)a + 5c + k
(25/2)a + 5c + k - (8a + 4c + k) = 65
4.5a + c = 65 **

when t = 9
s = (81/2)a + 9c + k
when t = 8
s = 32a + 8c + k
(81/2)a + 9c + k - (32a + 8c + k) = 105
8.5a + c = 105 ***

subtract ***-**
4a = 40
a = 10
then in **
c = 20
back in first:


s = 5t^2 + 20t + k

when t = 21
s = 2205 + 420 + k = 2625+k
when t = 20
s = 2000 + 400 + k = 2400+k

distance = 2625+k - 2400 - k = 225 metres

A physicist might have a better way of doing this.

Answer 3

v at 19.5 = v at 8.5 + 10*11
= 105 + 110
= 215 meters/second at 19.5
that is average speed from t = 19 to t = 20
so
215 meters