Instantaneous Acceleration To find instantaneous acceleration from the graph, calculate the slope of the graph at that instant of time. To do that, draw the tangent at that instant of time and calculate the slope of the tangent.
ainst at t=2s;
ainst= slope of the tangent =RunRise=t2−t1V2−V1=2−13.75−1.25= 2.5 m/s2
Note: for bodies moving with variable acceleration; ainst≠aavg
Displacement as the Area Under the Graph Using Velocity Vs. Time graph you can find the displacement by calculating the area under the graph.
Displacement from 1s to 5s;
Area of a triangle = 21 base × height = 21 × 40 × 4 = 80m
Note: Since the motion is along a straight line, the displacement is equal to distance covered.
Let’s consider the following Velocity Vs. Time graph, where the object changes direction as it moves.
Total displacement; area under the graph
Area of a triangle = 21 base × height = 21 × 10 × 50 = 250m
Total distance covered; distance covered from 0s to 5s + distance covered from 5s to 10s
Total distance = 50m + 50m = 100m
Note: Distance and Displacement are different, since the object changed direction.
A taxi driver travels across the city through the map as shown below. The total time taken to cover the whole journey is 30min.
A person walks 20m west, then 50m north and stops. After 5 minutes of resting, he resumes his walk towards the east for 20m. The entire trip took 10 minutes.
The following graph represents position Vs. time graph of a car as a function of time.
The following graph represents position Vs. time graph for an object as a function of time.
The following graph represent the Velocity Vs. Time graph of an object moving with constant acceleration.
The following represents the motion of a particle with a variable acceleration.
Define: Distance VS. displacement, speed VS. velocity, acceleration position, velocity, acceleration - time graphs
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