Three bodies A, B and Care moving in the same direction along a straig

1.	Three bodies A, B and Care moving in the same direction along a straight line path.From the graph of their motiona. Find their acceleration and arrange themin ascending order.b. Find the distance covered by A and B beforethey meet each other.
Answer» 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 brainly.in/question/11637622

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

Answer»

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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