Question

Three bodies A, B and Care moving in the same
direction along a straight line path.
From the graph of their motion
a. Find their acceleration and arrange them
in ascending order.
b. Find the distance covered by A and B before
they meet each other.​

Answer 1

Answer:

Part 1)

Acceleration of A = 2.5 m/s/s

acceleration of B = 10 m/s/s

acceleration of C = 0

so the ascending order of acceleration is

C < A < B

Part 2)

Distance covered by A is 240 m and distance covered by B is 80 m before they meet C

Explanation:

Part 1)

As we know that the acceleration of an object is rate of change in velocity

So here we will have

$a = \frac{dv}{dt}$

now we know that if the graph between velocity and time is given then rate of change in velocity is slope of the graph

So we have

acceleration of C = 0

acceleration of A = $\frac{40 - 20}{8 - 0}$

$a_A = 2.5 m/s^2$

acceleration of B = $\frac{40 - 0}{8 - 4}$

$a_B = 10 m/s^2$

so the order of acceleration of three are

C < A < B

Part B

Distance covered by B before it meet with C is given by the area under the curve

so we will have

$d_B = \frac{1}{2}(40)(8 - 4)$

$d_B = 80 m$

Distance covered by A before it meet with C is given by the area under the curve

so we will have

$d_A = \frac{1}{2}(40 + 20)(8 - 0)$

$d_A = 240 m$

#Learn

Topic : Kinematics

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