Question

study the given graph and answer the following questions from it:
(1) which part of the graph shows accelerated motion. Calculate the acceleration .
(2)which part of the graph shows retarded motion. Calculate the retardation .
(3)Calculate the distance of the body in first 4 seconds of the journey, graphically.

Answer 1

Given graph is a VELOCITY-TIME graph of an object.

In VT graph:
1. The slope of the line at a particular time gives the acceleration of the object at that time.

2. Area under the graph gives the displacement of the object.
_______________________________

Coming to the questions:

1. Part AB is accelerated motion as the slope of the graph(which gives the acceleration) is positive.

acceleration = slope

$slope = \frac{(4 - 0) \: m/s}{(4 - 0)s} = \frac{4}{4} = 1 \ m/s^2$

So acceleration = 1 m/s^2


2.Part CD is retarded motion as the slope of the graph(which gives the acceleration) is negative.

$slope = \frac{(0 - 4) \: m/s}{(14 - 10)s} = \frac{ - 4}{4} = - 1 \ m/s^2$

So acceleration = -1 m/s^2

Thus retardation is 1 m/s^2


3.
distance of the body in first 4 seconds = area under graph from t=0 to t=4s

area under graph from t=0 to t=4s = area of ∆ABE = 1/2 × 4 × 4 = 8m

Thus distance of the body in first 4 seconds is 8m

Answer 2

Answer:

From A to B - it is acceleration, C to D - retardation and distance travelled in first 4s is 8m.

Explanation:

Given graph is a velocity - time graph of an object.

1.  In v-t graph from A to B  has positive slope, hence this region there is   an accelerated motion

          acceleration is given by the slope of the curve

                            $a=\frac{\Delta y}{\Delta x}= \frac{4-0}{4-0} =1m/s^{2}$

        Thus, acceleration is $1 m/s^2$

2. From C to D, there is a negative slope, which shows the retarded motion.

           Slope of the curve is

                            $a=\frac{\Delta y}{\Delta x}= \frac{0-4}{14-12} =-1m/s^{2}$

         Thus, retardation is $-1 m/s^2$

3. Distance of the body in first 4 seconds is equal to the area under graph from t=0 to t=4s

           i.e., area of ∆ABE $=\frac{1}{2} b h=\frac{1}{2} \times 4\times 4 = 8m$

         Thus, distance travelled by the body in first 4 seconds is 8m