Solved Example Problems for Accelerated Motion
A velocity–time graph is given for a particle moving in x direction, as below
a) Describe the motion qualitatively in the interval 0 to 55 s .
b) Find the distance and displacement travelled from 0 s to 40 s .
c) Find the acceleration at t = 5 s and at t = 20 s
a) From O to A: (0 s to 10 s )
At t = 0 s the particle has zero velocity.
At t > 0, particle has positive velocity and moves in the positive x direction.
From 0 s to 10 s the slope (dv/dt) positive, implying the particle is accelerating. Thus the velocity increases during this time interval.
From A to B: (10 s to 15 s )
From 10 s to 15 s the velocity stays constant at 60 m s-1. The acceleration is 0 during this period. But the particle continues to travel in the positive x-direction.
From B to C : (15 s to 30 s )
From the 15 s to 30 s the slope is negative, implying the velocity is decreasing. But the particle is moving in the positive x direction. At t = 30 s the velocity becomes zero, and the particle comes to rest momentarily at t = 30 s .
From C to D: (30 s to 40 s )
From 30 s to 40 s the velocity is negative. It implies that the particle starts to move in the negative x direction. The magnitude of velocity increases to a maximum 40 m s-1
From D to E: (40 s to 55 s )
From 40 s to 55 s the velocity is still negative, but starts increasing from –40 m s-1 At t = 55 s the velocity of the particle is zero and particle comes to rest.
(b) The total area under the curve from 0 s to 40 s will give the displacement. Here the area from O to C represents motion along positive x–direction and the area under the graph from C to D represents the particle's motion along negative x–direction.
The displacement travelled by the particle from 0 s to 10 s = 1/2 × 10 × 60 = 300m
The displacement travelled from 10 s to 15 s = 60 × 5 = 300 m
The displacement travelled from 15 s to 30 s =1/2 × 15 × 60 = 450m
The displacement travelled from 30 s to 40 s = 1/2 × 10 × (-40) = -200m.
Here the negative sign implies that the particle travels 200 m in the negative x direction.
300 m + 300 m + 450 m − 200 m = +850 m.
Thus the particle's net displacement is along the positive x-direction.
The total distance travelled by the
particle from 0 s to 40 s = 300 + 300 + 450 + 200 = 1250 m.
(c) The acceleration is given by the slope in the velocity-time graph. In the first 10 seconds the velocity has constant slope (constant acceleration). It implies that the acceleration a is from v1 = 0 to v2 = 60 m s-1.
Next, the particle has constant negative slope from 15 s to 30 s . In this case v2=0 and v1=60m s-1. Thus the acceleration at t = 20 s is given by a = (0-60)/(30-15) = 4 m s-2. Here the negative sign implies that the particle has negative acceleration.
If the position vector of the particle is given by Find the
a. The velocity of the particle at t = 3 s
b. Speed of the particle at t = 3 s
c. acceleration of the particle at time t = 3 s
An object is thrown vertically downward. What is the acceleration experienced by the object?
We know that when the object falls towards the Earth, it experiences acceleration due to gravity g = 9.8 m s-2 downward. We can choose the coordinate system as shown in the figure.
The acceleration is along the negative y direction.