An automobile starting from rest at t0 = 0 at x0 = 0 undergoes constant acceleration in a straight line. It is observed to pass two marks separated by 64 m, the first at t = 8s and the second at t = 12s. What is the value of the acceleration?
To find the value of acceleration, we can use the following kinematic equation:
x = x0 + v0t + (1/2)at^2
Where:
x = final position
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Let's start by finding the initial velocity (v0) of the automobile.
Given:
x0 = 0 (starting from rest)
t0 = 0
Using the equation x = x0 + v0t + (1/2)at^2, we can find v0:
0 = 0 + v0(8) + (1/2)a(8)^2
0 = 8v0 + 32a ... (Equation 1)
Now, let's use the second mark at t = 12s to find the final position (x) of the automobile.
Given:
x1 = 64 m (position at t = 8s)
t1 = 8s
Using the equation x = x0 + v0t + (1/2)at^2, we can find x:
64 = 0 + v0(12) + (1/2)a(12)^2
64 = 12v0 + 72a ... (Equation 2)
Now, subtracting Equation 1 from Equation 2, we can eliminate v0:
64 - 0 = 12v0 - 8v0 + 72a - 32a
64 = 4v0 + 40a
Simplifying further:
64 = 4(v0 + 10a) ... (Equation 3)
From Equation 3, we can see that v0 + 10a = 16.
Now, substituting v0 = 16 - 10a into Equation 1:
0 = 8(16 - 10a) + 32a
0 = 128 - 80a + 32a
0 = 128 - 48a
48a = 128
a = 128/48
a = 2.67 m/s^2
Therefore, the value of acceleration is approximately 2.67 m/s^2.
To find the value of acceleration, we can use the equations of motion. Since the automobile is undergoing constant acceleration in a straight line, we can use the following equation:
x = x0 + v0t + (1/2)at^2,
where x is the position of the automobile at time t, x0 is the initial position (which is 0 in this case), v0 is the initial velocity (which is 0 since the automobile starts from rest), a is the acceleration, and t is the time.
We are given that the automobile passes two marks separated by 64 m, the first mark at t = 8s and the second mark at t = 12s. Let's use this information to set up two equations:
For the first mark (t = 8s):
x1 = x0 + v0(8) + (1/2)a(8)^2
For the second mark (t = 12s):
x2 = x0 + v0(12) + (1/2)a(12)^2
Since x0 is 0 and v0 is 0, these equations simplify to:
x1 = (1/2)a(8)^2
x2 = (1/2)a(12)^2
We are given that the distance between the two marks is 64 m, so we can set up another equation:
x2 - x1 = 64
Substituting the expressions for x1 and x2, we get:
(1/2)a(12)^2 - (1/2)a(8)^2 = 64
Simplifying this equation, we get:
36a - 16a = 64
20a = 64
a = 64/20
a = 3.2 m/s^2
Therefore, the value of the acceleration is 3.2 m/s^2.
d = 1/2 a t^2
1/2 a(12^2 - 8^2) = 64