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
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
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.
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
To find the distance traveled by the body in the 20th second, we need to understand the pattern of motion. From the given information, we know that the body is accelerating uniformly.
First, let's find the acceleration of the body. We can use the formula for displacement (s) in terms of time (t) and acceleration (a) when the initial velocity (u) is zero:
s = ut + (1/2)at^2
Since the initial velocity is zero, the equation simplifies to:
s = (1/2)at^2
Using the information given, we can find the acceleration:
65 = (1/2)a(5^2)
130 = 25a
a = 130/25
a = 5.2 m/s^2
Now that we have the acceleration, we can find the distance traveled in the 9th second using the same equation:
105 = (1/2)(5.2)(9^2)
105 = (1/2)(5.2)(81)
105 = 211.68
The value does not match. Let's return and find the true acceleration.
To find the true acceleration, we can use the formula for average acceleration:
a = (v - u) / t
where "v" is the final velocity, "u" is the initial velocity, and "t" is the time interval.
In this case, the initial velocity is zero, and we know the distances traveled in the 5th and 9th seconds. So:
a = (105 - 65) / (9 - 5)
a = 40 / 4
a = 10 m/s^2
Now that we have the correct acceleration, we can proceed to find the distance traveled in the 20th second.
Using the same equation as before:
s = (1/2)at^2
s = (1/2)(10)(20^2)
s = (1/2)(10)(400)
s = 2000 meters
Therefore, the body will travel 2000 meters in the 20th second.