We consider the s-t graph of an object that moves in a straight line. In the s-t graph, velocity = slope of the line at a given point on the s-t graph. OA = the object moves at a steady speed because its slope is constant. The slope of a straight line is constant. It does not change with time. There are so many examples. One example is a car moves at a steady speed on a straight road.
In the OCB graph, Its velocity is changing always with time because the slope of the graph is changing with time always. The velocity at the time t1 = slope or gradient at the point C on the graph.
How can we find the gradient at the point on the graph
First, we should choose a point on the graph. Next, we plot an antenna on the point which graph and line t1C are intersected. Select two points as far away as possible on the antenna. Now we calculate the slope by using this formula. More information about the graph
The gradient of the s-t graph = velocity
Velocity – time graph
The velocity at the time t1 = slope or gradient at the point F on the graph as shown in this fig. In this fig, you can see 3 types of v-t graphs.
They are DE, OA, and OFCB.
OA line has a constant gradient. Therefore it moves constant acceleration. The acceleration = gradient of the v-t graph. A body starts from rest and with uniform acceleration. Its velocity is continuously increasing with time. There are so many examples. One is a free-falling body under gravity.
It has constant velocity. Or the body is moving at a steady speed. Every time its acceleration is zero.
Its gradient or slope changes with time. This body is started with rest because the graph is started from the point of (0,0). Its mean velocity = 0, time = 0. While we are seeing from point O to F on the blue-colored line, the slope of the graph is decreased from a positive value to a minimum value. You can see it well. It means the body is decreased its velocity from V0 to V1 within time t1. Here, t1 is in seconds. V0 > V1 . From F to C, the slope of the graph is increased. It means the velocity of the body is increased from V1 to V2 within (t2 – t1 ) seconds. Let’s assume the velocities of the body at the points O, F, and C respectively V0, V1, and V2. The acceleration of this motion is changed with time. It is positive from O to B. After B the gradient decreases from a positive value to zero when the slope exactly parallel to the t-axis. After that gradient decreases from zero to negative.
The distance body has moved within the time of (t2 – t1 ) seconds = Average speed * Time
The area between a velocity graph and t-axis represents the displacement of the body during the time.
A body starts with an initial velocity of u and is moving in a straight line with a constant acceleration of a. After a time of t, let’s assume its velocity is v and its distance from the start is s. We can show this case in a v-t graph.
We can find the acceleration by the gradient of the graph. The gradient of the graph is tanθ. lets us find it,
, The distance = area between the graph and t-axis
we can get the other equations also.
An Example of Motions in a straight line.
- A train travels in a straight line at a constant acceleration of 2 ms-2. The train is increased its speed from 10 ms-1 to v ms-1 within 30 s.
- Calculate the v by using a v-t graph of this motion.
- Draw the s-t graph and a-t graph.
acceleration = gradient of the v-t graph
v-t graph = gradient of the s-t graph. Therefore we can draw the s-t graph according to its v-t.
We can find the distance also between the velocities of 10 and 70.
It has traveled 900m in 30 seconds.
Drawing the s-t graph:
First, sketch down the same two axes below the v-t graph as shown in this fig. Rename its two axes to the s – t. Now we find the value of Point A on the graph s-t according to Point A on the graph v-t. The v-axis value of point A on the graph v-t (10)=The gradient of the s-t graph at point A=.
The v-axis value of point B on the graph v-t (70)=The gradient of the s-t graph at point B=. . Therefore we can decide how the curve should be plotted from A to B. It should increase its gradient or angle from A to B. You can see the orange-colored line in fig.
Drawing the a-t graph:
Now we want to draw the a-t graph from the v-t graph. Same as the above first we want to sketch down the two axes and name its axis as a and t as shown in fig. Now we want to find the initial value, end value and how is the gradient or angle of the s-t graph between t=0 and t=30. We know already that, the gradient of the v-t graph = a-t graph.
In this case, we can see the gradient(2) of the v-t graph is constant and positive all the time. It is simple. You can see the orange-colored straight line in the fig.
More examples of linear motions.
- Choose the right expressions through the following expressions.
- An Object which travels with constant acceleration can have a zero velocity
- The velocity of an object which moves at a steady speed can be changed with time.
- The speed of an object which travels at a constant velocity can be changed with time.
- An object accelerates to the West, has a velocity to the East.
- The direction of the object which travels with constant acceleration can be changed.
- A motor bicycle that travels at a speed of 20 ms-1 is braked and stopped under the deacceleration of 1 ms-2. Find the distance and the time which travels with deacceleration.
- A train that starts from rest at a station is traveled its first 2km at a constant acceleration and next, it is traveled 6km at a steady speed. Finally, it is traveled 1km with constant deacceleration until it comes to the rest. If the distance between the two stations is 12km and the time it has taken is 20minutes find the
- acceleration of the train by using a v-t graph.
- steady speed.
- Find the maximum height and flying time of the throw of a small stone which is thrown vertically upward at an initial speed of 40 ms-1. Consider g = 10 ms-2 .
- A coconut falls from a tree to the ground within 4s. Find
- the height of the tree.
- the speed of the coconut just before impact with the ground.
- Find the flying time of a stone which is thrown vertically upward at an initial speed of 25 ms-1 from atop of a 50m high tower.
- The s-t graph of the object which travels to the x-direction is showing in the following fig. The acceleration or de-acceleration of the object is constant. Draw
- v-t graph
- a-t graph for this motion.
- Given below a-t graph indicates the way of changing the acceleration(a) of an object with the time(t). Drow a
- v-t graph
- s-t graph for this motion.
- Find the resultant velocity of an object which 10 ms-1 and 12 ms-1 are acted perpendicularly on each other on the object.
- Two velocities of 8 ms-1 and 6 ms-1 are acted on an object with an angle of 450. Find the resultant velocity of the object.