A body moving with uniform acceleration. covers 65 centimeters in the fifth second and 105 centimeter in the ninth second. What distance? Does it travel in 15 seconds?
To find the distance traveled in 15 seconds, we need to find the acceleration of the body.
The formula to calculate distance covered with uniform acceleration is:
Distance = Initial velocity × Time + (1/2) × Acceleration × Time²
We are given the distance covered in the 5th second and the 9th second. Using these values, we can set up two equations:
65 = Initial velocity × 5 + (1/2) × Acceleration × (5)²
105 = Initial velocity × 9 + (1/2) × Acceleration × (9)²
From the first equation, we can rewrite it as:
65 = 5Initial velocity + 12.5Acceleration
From the second equation, we can rewrite it as:
105 = 9Initial velocity + 40.5Acceleration
Now, we have a system of two equations with two variables (Initial velocity and Acceleration).
To solve it, we can use the method of substitution by rewriting the first equation to isolate the Initial velocity:
5Initial velocity = 65 - 12.5Acceleration
Initial velocity = (65 - 12.5Acceleration)/5
Now, substitute this value of Initial velocity into the second equation:
105 = 9(65 - 12.5Acceleration)/5 + 40.5Acceleration
Multiply throughout by 5 to remove the denominator and simplify:
525 = 9(65 - 12.5Acceleration) + 202.5Acceleration
525 = 585 - 112.5Acceleration + 202.5Acceleration
525 = 585 + 90Acceleration
90Acceleration = 585 - 525
90Acceleration = 60
Acceleration = 60/90
Acceleration = 2/3 cm/s²
Now, substitute the value of Acceleration back into one of the original equations to solve for Initial velocity:
65 = 5Initial velocity + 12.5(2/3)
65 = 5Initial velocity + 25/3
(3/3)(65) = (3/3)(5Initial velocity) + 25/3
195/3 - 25/3 = 5Initial velocity
(195 - 25)/3 = 5Initial velocity
170/3 = 5Initial velocity
Initial velocity = 170/15
Initial velocity = 34/3 cm/s
Now that we have the Initial velocity (34/3 cm/s), we can use it to find the distance covered in 15 seconds using the formula:
Distance = Initial velocity × Time + (1/2) × Acceleration × Time²
Distance = (34/3) × 15 + (1/2) × (2/3) × (15)²
Distance = (34/3) × 15 + (1/2) × (30/3) × (225)
Distance = 510/3 + 15 × 75
Distance = 170 + 1125
Distance = 1295 cm
Therefore, the body will travel a distance of 1295 centimeters in 15 seconds.
To find the distance traveled in 15 seconds, we need to find the acceleration first.
In the fifth second, the body covers 65 centimeters.
Using the formula for distance traveled with uniform acceleration:
\[d = ut + \frac{1}{2}at^2\]
where d is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Since the body starts from rest (initial velocity u = 0), this simplifies the formula to:
\[d = \frac{1}{2}at^2\]
Given that it covers 65 centimeters in the fifth second, we can substitute the values into the formula:
\[65 = \frac{1}{2}a(5)^2\]
Solving for 'a':
\[65 = \frac{25}{2}a\]
\[a = \frac{65 \times 2}{25}\]
\[a = \frac{130}{25}\]
\[a = 5.2\]
So, the acceleration is 5.2 centimeters per second squared.
Now, using the same formula, we can find the distance traveled in the ninth second.
\[d = \frac{1}{2}a(9)^2\]
\[105 = \frac{1}{2}a(81)\]
Solving for 'a':
\[105 = \frac{81}{2}a\]
\[a = \frac{105 \times 2}{81}\]
\[a = \frac{210}{81}\]
\[a ≈ 2.59\]
So, the acceleration is approximately 2.59 centimeters per second squared.
Now that we have the acceleration, we can find the distance traveled in 15 seconds using the formula:
\[d = \frac{1}{2}at^2\]
\[d = \frac{1}{2}(2.59)(15)^2\]
Calculating:
\[d ≈ \frac{1}{2}(2.59)(225)\]
\[d ≈ 291.975\]
Therefore, the body will travel approximately 291.975 centimeters in 15 seconds.