Wiki User
∙ 11y agoit measures the magnitude of acceleration, but it can't tell you the direction of the acceleration.
Wiki User
∙ 11y agoThe slope of the tangent to the curve on a velocity-time graph represents the acceleration of an object. Positive slope indicates acceleration in the positive direction, negative slope indicates acceleration in the negative direction, and zero slope indicates constant velocity.
The slope of a tangent to the curve of a velocity-time graph represents the acceleration of an object at that specific instant in time. A steeper slope indicates a greater acceleration, while a flatter slope indicates a smaller acceleration.
The instantaneous speed at a specific point on a speed-time graph is the slope of the tangent to the curve at that point. It represents the speed of an object at that exact moment in time. This can be determined by calculating the gradient at that particular point.
To obtain average velocity from a displacement-time graph, divide the total displacement by the total time taken. For instantaneous velocity, find the slope of the tangent to the curve at a specific point on the graph. This tangent represents the velocity at that instant.
To obtain the average velocity from a displacement-time graph, you can calculate the slope of the line connecting two points on the graph. Divide the change in displacement by the change in time. To obtain the instantaneous velocity, you need to find the slope of the tangent line at a specific point on the graph. Choose a point on the graph and draw a line tangent to the curve at that point. The slope of this tangent line will give you the instantaneous velocity at that specific point.
Acceleration on a speed-time graph is represented by the slope of the line. A steeper slope indicates a greater acceleration. If the slope is positive, it indicates acceleration; if it is negative, it indicates deceleration.
You find the slope of the tangent to the curve at the point of interest.
The slope of a tangent to the curve of a velocity-time graph represents the acceleration of an object at that specific instant in time. A steeper slope indicates a greater acceleration, while a flatter slope indicates a smaller acceleration.
No, average velocity is the total displacement divided by the total time taken. The slope of the tangent to the curve on a velocity-time graph at a specific instant of time gives the instantaneous velocity at that moment, not the average velocity.
Tangent line is a graph. This graph is to gather data.
A tangent to a velocity-time graph represents the instantaneous acceleration of an object at that specific moment in time. It shows how the velocity is changing at that particular point.
The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.
Slope = (vertical change)/(horizontal change), commonly referred to as rise/run. If the graph is a straight line, then you can count squares or measure how much change in vertical, over a specified change in horizontal. If it is a curve, then you need to have a tangent line (a line that touches the curve at a specific point and has the same slope as the line), then you can determine the slope of that line using the method described, above.
To find instantaneous velocity from a position-time graph, you calculate the slope of the tangent line at a specific point on the graph. The slope represents the rate of change of position at that instant, which is equivalent to the velocity at that particular moment.
The variable plotted along the vertical axis is the distance in the first case, speed in the second. The gradient of (the tangent to) the distance-time graph is the speed while the area under the curve of the speed-time graph is the distance.
When you graph a tangent function, the asymptotes represent x values 90 and 270.
Draw a tangent to the curve at the point where you need the gradient and find the gradient of the line by using gradient = up divided by across
From v = u - (a*t)then:a = (v-u) / tSelect a point on a time(x) - velocity(y) graph, calculate the slope of a tangent to the line at this point , this is the rate of acceleration at that point.On a data driven curve approximation is used.On a mathematical curve differential calculus can be used.