A body moving with constant acceleration travels the distance 6 m and 16 m respectively in 2 s and 4 s. Calculate the initial velocity and acceleration of the body

Question

A body moving with constant acceleration travels the distance 6 m and 16 m respectively in 2 s and 4 s. Calculate the initial velocity and acceleration of the body

Answer 1

let the acceleration be a m/s^2

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

at the beginning of our observation:
t = 0 , s = 0 , then k = 0

s = (a/2) t^2 + ct

when t = 2
6 = (a/2)(4) + 2c
6 = 2a + 2c
a+c = 3

when t = 4
16 = (a/2)(16) + 4c
16 = 8a + 4c
4 = 2a + c

subtract those last two equations:
c = 2
and then a = 1

the acceleration is 1 m/s^2
when t = 0 (initial)
v = at + c
= 1(0) + 2
= 2

and the initial velocity is 2 m/s

I did this using Calculus. They have formulas for this in
physics, perhaps using them would be shorter and simpler.

Answer 2

To find the initial velocity and acceleration of the body, we can use the equations of motion for uniformly accelerated motion.

The distance traveled by the body is given as 6 m in 2 s and 16 m in 4 s.

Let's denote the initial velocity by u and the acceleration by a.

Using the equation for distance traveled in terms of initial velocity, acceleration, and time:

Distance = initial velocity * time + (1/2) * acceleration * time^2

For the first case, when the body travels a distance of 6 m in 2 s:

6 = u * 2 + (1/2) * a * 2^2
6 = 2u + 2a

For the second case, when the body travels a distance of 16 m in 4 s:

16 = u * 4 + (1/2) * a * 4^2
16 = 4u + 8a

We now have a system of two equations with two variables:
6 = 2u + 2a
16 = 4u + 8a

We can solve this system of equations to find the values of u and a.

Multiplying the first equation by 4, we get:
24 = 8u + 8a

Subtracting this equation from the second equation, we have:
16 - 24 = 4u + 8a - (8u + 8a)
-8 = 4u - 8u
-8 = -4u
u = -8 / -4
u = 2 m/s

Substituting the value of u back into the first equation, we find:
6 = 2 * 2 + 2a
6 = 4 + 2a
2a = 6 - 4
2a = 2
a = 2/2
a = 1 m/s^2

Therefore, the initial velocity of the body is 2 m/s and the acceleration is 1 m/s^2.

Answer 3

To calculate the initial velocity and acceleration of a body moving with constant acceleration, we can use the following equations:

1. The formula for distance traveled by an object with constant acceleration is given by:

𝑠 = 𝑢𝑡 + 0.5𝑎𝑡^2

where 𝑢 is the initial velocity, 𝑡 is the time taken, 𝑎 is the acceleration, and 𝑠 is the distance traveled.

We can use this equation to form two equations using the given information.

For the first distance of 6 m traveled in 2 s:
6 = 𝑢(2) + 0.5𝑎(2)^2

For the second distance of 16 m traveled in 4 s:
16 = 𝑢(4) + 0.5𝑎(4)^2

2. We have two equations with two unknowns (𝑢 and 𝑎). We can solve these equations simultaneously to find the values of 𝑢 and 𝑎.

From the first equation:
6 = 2𝑢 + 2𝑎

From the second equation:
16 = 4𝑢 + 8𝑎

3. Rearranging the equations, we get:

2𝑢 + 2𝑎 = 6 ---(Equation 1)
4𝑢 + 8𝑎 = 16 ---(Equation 2)

4. We can now solve the simultaneous equations. One way to do this is to use the method of substitution.

From Equation 1, we can isolate 𝑢 in terms of 𝑎:
𝑢 = (6 - 2𝑎) / 2

5. Substituting the value of 𝑢 in Equation 2, we have:

4((6 - 2𝑎) / 2) + 8𝑎 = 16

Simplifying this equation:

12 - 4𝑎 + 8𝑎 = 16

4𝑎 = 4

𝑎 = 1 m/s^2

6. Now that we know the acceleration (𝑎 = 1 m/s^2), we can substitute this value back into Equation 1 to find 𝑢:

2𝑢 + 2(1) = 6

2𝑢 + 2 = 6

2𝑢 = 6 - 2

2𝑢 = 4

𝑢 = 2 m/s

7. Finally, we have determined that the initial velocity (𝑢) is 2 m/s and the acceleration (𝑎) is 1 m/s^2 for the given scenario.