1.Moving with uniform acceleration a body covers 150m during 10sec so thet it covers 24m during the 10th sec .What is the initial velocity and the acceleration of the body.
2.A runner travels 1.5 laps around a circular track of 50sec .The diameter of the track is 40m and its circumferenceis 126m.Find a)average speed of the runner and b)the magnitude of the runner average velociy
Someone will gladly critique your thinking, when any is shown.
1. To find the initial velocity and acceleration of the body, we can use the following equations of motion:
Distance covered (S) = Initial velocity (u) * Time (t) + (1/2) * Acceleration (a) * Time^2
Given:
Total distance covered (S) = 150 m
Time (t) = 10 s
Distance covered in the 10th second = 24 m
From the equation above, we can write two equations using the given values:
1) 150 = u * 10 + (1/2) * a * (10)^2
2) 24 = u * 9 + (1/2) * a * (9)^2
We have two equations with two variables (u and a).
Solving these equations simultaneously will give us the values of initial velocity (u) and acceleration (a).
2. To find the average speed of the runner, we can use the formula:
Average speed (S) = Total distance covered / Total time taken
Given values:
Total distance covered = 1.5 laps * circumference of the track = 1.5 * 126 m
Total time taken = 50 s
Substituting these values in the formula will give us the average speed.
To find the magnitude of the runner's average velocity, we can divide the displacement by the total time taken:
Average velocity (V) = Displacement / Total time taken
Since the runner is making laps around the circular track, the displacement is 0.
Therefore, the magnitude of the runner's average velocity is 0 m/s.
1. To find the initial velocity and acceleration of the body, we can use the equations of motion.
First, we know that the body covers a distance of 150m in 10 seconds. Let's label this as the total distance traveled, D = 150m, and the total time, T = 10s.
During the 10th second, the body covers an additional distance of 24m. This means that the body covered 150m - 24m = 126m in the first 9 seconds. Let's label this as the distance covered in the first 9 seconds, d1 = 126m.
Using the equation of motion for distance covered with uniform acceleration:
D = (1/2)at^2 + v0t
where D is the total distance covered, a is the acceleration, t is the time, and v0 is the initial velocity.
Substituting the given values into the equation, we have:
150m = (1/2)a(10s)^2 + v0(10s) (equation 1)
Next, we can use the information about the distance covered during the 10th second to find the acceleration.
During the 10th second, the body covers a distance of 24m. We can use the equation of motion for distance covered with constant velocity:
d2 = v0 + v^2 / (2a)
where d2 is the distance covered during the 10th second, v0 is the initial velocity, v is the final velocity, and a is the acceleration.
Substituting the known values, we have:
24m = v0 + v^2 / (2a) (equation 2)
Now we have a system of equations with two unknowns (v0 and a), and we can solve it by substituting equation 2 into equation 1:
150m = (1/2)a(10s)^2 + (v0 + v^2 / (2a))(10s)
Simplifying the equation, we get:
150m = 50a + 10v0 + (v0 + v^2 / a)
Expanding further, we have:
150m = 60v0 + 50a + v^2 / a
From equation 2, we can replace v0 + v^2 / (2a) with 24m:
150m = 60(24m) + 50a + v^2 / a
Simplifying, we get:
150m = 1440m + 50a + v^2 / a
Rearranging the equation, we have:
v^2 / a = 150m - 1440m - 50a
v^2 / a = -1290m - 50a
Taking the derivative with respect to a, we have:
2v(dv/da) - v^2 / a^2 = -50 - 50(da/da)
=> 2v(dv/da) - v^2 / a^2 = -50 - 50
Simplifying further, we get:
2v(dv/da) - v^2 / a^2 = -100
Now, we can solve this equation to find the value of v. Once we have v, we can substitute it back into equation 2 to find v0.
2. To find the average speed of the runner, we need to divide the total distance traveled by the total time taken.
Given that the runner travels 1.5 laps around a circular track, we can calculate the total distance covered by multiplying the circumference of the track by 1.5:
Total distance = 1.5 x Circumference = 1.5 x 126m
To find the average speed, we divide the total distance by the total time:
Average speed = Total distance / Total time = (1.5 x 126m) / 50s
To find the magnitude of the runner's average velocity, we need to divide the displacement by the total time:
Average velocity = Displacement / Total time
However, since the runner is moving in a circular track, the displacement is zero, and therefore the magnitude of the average velocity is also zero. This is because velocity is a vector quantity, and displacement measures the change in position of an object. Since the runner ends at the same position where they started on a circular track, the displacement is zero.
1.
s =vₒ•t +a•t²/2,
v(9) = vₒ +a•9,
24 =v(9) •1 + a•1²/2 =
=vₒ +a•9 + a•1²/2 =
= vₒ +9.5•a.
s =vₒ•t +a•t²/2 =>
150 = vₒ•10+ a•100/2 =
= 10• vₒ + 50•a.
24= vₒ +9.5•a,
150= 10• vₒ + 50•a.
45° = 90,
a = 2 m/s².
vₒ = 24-9.5 •a = 5 m/s.
2.
Average speed = total distance travelled/ total time taken = 1.5•126/50 = 3.78 m/s,
Displacement is the distance from the starting point to the end point.
Average velocity = final displacement/total time taken = D/t = =40/50 =0.8 m/s.