Question

Study the given graph and find the distance covered by the object between A to D.

67.5 m
76.5 m
46 m
46m/s​

Answer 1

Answer:

hope it helps you

please marks as brainliest

Explanation:

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^2slope=

(4−0)s

(4−0)m/s

=

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^2slope=

(14−10)s

(0−4)m/s

=

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:

s

d=s*t

so area covered by graph will be the distance