A body covers a distance of 25m in the 5th sec of its motion and 40m in the 8th sec.of its motion. Calc.initial velocity and acceleration.
calculate velocity in the fifth second and the 8th second.
veloicty(t=5)=25m/1sec
velocity(t=8)=40m/s
acceleration=changevelocity/time=15m/s/(3sec)
=5m/s^2
To calculate the initial velocity and acceleration, we can use the equations of motion:
1. The equation for distance traveled during constant acceleration is:
s = ut + (1/2)at^2
2. The equation for velocity is:
v = u + at
Using the given information, we can create two equations:
From the first equation, when the body covers a distance of 25m in the 5th second:
25 = u(5) + (1/2)a(5)^2 --> Equation 1
From the second equation, when the body covers a distance of 40m in the 8th second:
40 = u(8) + (1/2)a(8)^2 --> Equation 2
To solve these equations simultaneously, we can eliminate the variable "u" by subtracting Equation 1 from Equation 2:
40 - 25 = u(8) + (1/2)a(8)^2 - u(5) - (1/2)a(5)^2
15 = 3u + (1/2)a(64) - 5u - (1/2)a(25)
15 = -2u + 32a - (25/2)a
To simplify the equation further, let's combine like terms:
15 = -2u + (64 - 25/2)a
15 = -2u + (128/2 - 25/2)a
15 = -2u + (103/2)a
Now, we can rearrange the equation to express "u" in terms of "a":
2u = (103/2)a - 15
u = [(103/2)a - 15]/2
Now, we have an expression for "u" in terms of "a". Let's substitute this expression into either Equation 1 or Equation 2 to solve for "a".
Let's substitute into Equation 1:
25 = [(103/2)a - 15]/2 * (5) + (1/2)a(5)^2
Multiplying through by 2:
50 = (103a - 30)(5) + 5a^2
Expanding:
50 = 515a - 150 + 5a^2
Rearranging the equation by moving all terms to one side:
5a^2 - 515a + 200 = 0
Now we can solve this quadratic equation for "a". Applying the quadratic formula:
a = (-(-515) ± sqrt((-515)^2 - 4(5)(200)))/(2*5)
= (515 ± sqrt(265225 - 4000))/10
= (515 ± sqrt(261225))/10
= (515 ± 511)/10
Therefore, the two possible values for "a" are:
1. a = (515 + 511)/10 = 1026/10 = 102.6 m/s^2
2. a = (515 - 511)/10 = 4/10 = 0.4 m/s^2
Now that we have the possible values for "a", we can substitute them back into the expression for "u" to find the corresponding values of the initial velocity "u".
For a = 102.6 m/s^2:
u = [(103/2)(102.6) - 15]/2
= (10589.05 - 15)/2
= 5287.05/2
= 2643.525 m/s
For a = 0.4 m/s^2:
u = [(103/2)(0.4) - 15]/2
= (20.6 - 15)/2
= 5.6/2
= 2.8 m/s
Therefore, the initial velocity can be either 2643.525 m/s or 2.8 m/s, depending on the value of acceleration.
To calculate the initial velocity and acceleration of the body, we need to use the equations of motion. Let's break down the problem step by step:
Step 1: Identify the given information:
- Distance covered in the 5th second = 25m
- Distance covered in the 8th second = 40m
Step 2: Determine the time interval:
The time interval between the 5th second and the 8th second is 8 - 5 = 3 seconds.
Step 3: Calculate the average velocity:
Average velocity is given by the formula:
Average Velocity (Vavg) = Total Distance / Total Time
The total distance covered between the 5th and 8th seconds can be calculated by subtracting the distance covered in the 5th second from the distance covered in the 8th second:
Total Distance = Distance covered in 8th second – Distance covered in 5th second
= 40m – 25m
= 15m
Therefore, Average Velocity = 15m / 3s = 5m/s
Step 4: Calculate the initial velocity and acceleration:
We can use the first equation of motion to find the initial velocity (u) using the formula:
V = u + at,
where
V = Final velocity = Average velocity = 5m/s (from step 3)
u = Initial velocity (to be determined)
a = Acceleration (to be determined)
t = Time interval between the 5th and 8th seconds = 3s
Substituting the given values into the equation, we get:
5 = u + a * 3
Similarly, we can use the second equation of motion to find the displacement (S) using the formula:
S = ut + (1/2) * a * t^2
where
S = Displacement = Total distance = 15m (from step 3)
Substituting the given values into the equation, we get:
15 = u * 3 + (1/2) * a * (3)^2
Now, we have two equations with two unknowns (u and a).
Solving the two equations simultaneously will give us the values of u and a.
The solution to these equations is:
u = 10m/s (Initial velocity)
a = -5m/s² (Acceleration)
Therefore, the initial velocity of the body is 10m/s, and the acceleration is -5m/s². The negative sign indicates deceleration or retardation in this case.