hope im right...
dx/dt = 3.1 t
dx = 3.1 t dt
dx=52
z=distance
52= intergral(0,Z) 3.1 t dt
52=3.1 Z^2 /2
Z^2=104/3.1
So it's acceleration is 4m/s2. So at any point because it says uniformly, it will be accelerating at 4m/s2 each second
Since the car accelerates uniformly, its average speed during the interval is1/2 (initial speed + final speed) = 1/2 (4.15 + 17.11) = 10.63 m/s.The distance covered is (average speed) x (time) = (10.63) x (4.8) = 51.024 meters.If the motion during the 4.8 seconds was all in the same direction along a straight line,then that same 51.024 meters is also the displacement during that time.If, instead, it happened to take 4.8 seconds to go around a circle with a circumferenceof 51.024 m, then the displacement over the 4.8 seconds is zero, since the starting lineis also the finish line.
To find the distance traveled by a point on the edge of the wheel, we first need to calculate the average angular velocity. The wheel accelerates from 240 rpm to 360 rpm, so the average angular velocity is (240 + 360) / 2 = 300 rpm. Converting this to radians per second, we have 300 rpm × (2π rad / 1 min) × (1 min / 60 s) = 31.42 rad/s. The wheel travels for 6.5 seconds, so the total angular displacement is 31.42 rad/s × 6.5 s = 204.23 radians. The circumference of the wheel is π × diameter = π × 0.33 m ≈ 1.04 m. Therefore, the distance traveled is 204.23 radians × 0.33 m/radian ≈ 67.39 m.
An arc, which is a uniformly curved line with a common radii point.
When the rate of change varies from point to point, the relationship is considered non-linear. In such cases, the change in one variable does not correspond to a constant change in another variable, resulting in a curve rather than a straight line on a graph. This can be observed in many real-world scenarios, such as exponential growth or decay, where the rate of change accelerates or decelerates over time.
So it's acceleration is 4m/s2. So at any point because it says uniformly, it will be accelerating at 4m/s2 each second
Since the car accelerates uniformly, its average speed during the interval is1/2 (initial speed + final speed) = 1/2 (4.15 + 17.11) = 10.63 m/s.The distance covered is (average speed) x (time) = (10.63) x (4.8) = 51.024 meters.If the motion during the 4.8 seconds was all in the same direction along a straight line,then that same 51.024 meters is also the displacement during that time.If, instead, it happened to take 4.8 seconds to go around a circle with a circumferenceof 51.024 m, then the displacement over the 4.8 seconds is zero, since the starting lineis also the finish line.
The displacement is 6km because it is the straight-line distance from the starting point to the final point. The displacement represents the shortest path between the two points, regardless of the actual distance traveled.
loads are carried out as point load uniformly distributed load and uniformly varying load
The furthest point I have traveled from home is approximately 5,000 miles away.
The distance moved by an object is the total length of the path traveled, while the displacement vector is the shortest distance between the starting and ending points in a straight line. Therefore, the distance moved can be greater than or equal to the displacement vector, depending on the path taken by the object.
A straight line is usually faster than a zigzag route when traveling from one point to another. Zigzagging increases the overall distance traveled and can result in a longer travel time compared to a straight path.
To find the distance traveled by a point on the edge of the wheel, we first need to calculate the average angular velocity. The wheel accelerates from 240 rpm to 360 rpm, so the average angular velocity is (240 + 360) / 2 = 300 rpm. Converting this to radians per second, we have 300 rpm × (2π rad / 1 min) × (1 min / 60 s) = 31.42 rad/s. The wheel travels for 6.5 seconds, so the total angular displacement is 31.42 rad/s × 6.5 s = 204.23 radians. The circumference of the wheel is π × diameter = π × 0.33 m ≈ 1.04 m. Therefore, the distance traveled is 204.23 radians × 0.33 m/radian ≈ 67.39 m.
distance from point x to point y, not the distance that traveled.
A displacement of 4 km from the starting point.
An arc, which is a uniformly curved line with a common radii point.
The slope at each point on a speed/time graph is themagnitude of acceleration at that point in time.