Solution of:An air plane ABC R=250m (x=30)
the speed of the meachine is 450Km/h in A determine the accelartion supported by the pilot in A and the tangential accelartion is 10m.s
determine the speed in B if the tangential accelaration is constant
To solve this problem, we can use the formula for acceleration:
a = Δv / Δt
Where
a = acceleration,
Δv = change in velocity, and
Δt = change in time.
Now, let's break down the problem into smaller parts:
1. Determining the acceleration supported by the pilot in A:
We are given that the tangential acceleration (a_t) is equal to 10 m/s^2. However, we also need to find the magnitude of the centripetal acceleration (a_c), so we can calculate the total acceleration.
The centripetal acceleration is given by the formula:
a_c = v^2 / R
Where
a_c = centripetal acceleration,
v = velocity, and
R = radius of curvature.
Here, R is given as 250m. To find the velocity (v), we need to convert the speed given in kilometers per hour (km/h) to meters per second (m/s). We can use the formula:
v = Speed / 3.6
In this case, the speed is given as 450 km/h. So,
v = 450 km/h / 3.6 = 125 m/s
Now we can find the centripetal acceleration:
a_c = (125 m/s)^2 / 250m = 62.5 m/s^2
To find the total acceleration, we add the tangential and centripetal acceleration:
a = a_t + a_c = 10 m/s^2 + 62.5 m/s^2 = 72.5 m/s^2
Therefore, the acceleration supported by the pilot in A is 72.5 m/s^2.
2. Determining the speed in B:
Since the tangential acceleration is constant, we can use the formula for velocity:
v = u + a_t * t
Where
v = final velocity,
u = initial velocity,
a_t = tangential acceleration, and
t = time.
In this case, we are looking for the final velocity (v) in point B when the tangential acceleration (a_t) is still 10 m/s^2.
Given that the initial velocity (u) is 125 m/s (from point A) and the tangential acceleration (a_t) is 10 m/s^2, we need to find the time it takes to travel from point A to point B.
We can use the kinematic equation:
s = u * t + (1/2) * a_t * t^2
Where
s = distance.
The distance (s) is given as x = 30m.
So,
30 = 125 * t + (1/2) * 10 * t^2
Rearranging the equation, we have a quadratic equation:
5t^2 + 125t - 60 = 0
We can solve this quadratic equation using the quadratic formula. Once we find the value of 't', we can substitute it back into the velocity equation to find the final velocity (v).
Please note that this process involves solving a quadratic equation, and the quadratic formula can sometimes yield two solutions. So, we'll need to determine which solution is physically meaningful in this context.
Once you find the value of 't', you can substitute it back into the velocity equation to find the final velocity (v).
By following these steps, you can determine the acceleration supported by the pilot in A and the speed in B if the tangential acceleration is constant.