Question

please solve this step by step
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Answer 1

Answer:

Knowledge required:

• In a velocity-time graph, the slope of graph tell us about the acceleration. If the slope is high then the acceleration is positive, if the solve is low then the acceleration is negative and if the slope is parallel to time axis then there is no acceleration taking place!

• In the velocity time graph the distance or displacement can be founded by the area under the curve!

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Question 6.

(i) Study the velocity time graph and calculate:

(a) The acceleration from A to B.

→ Here, the final velocity is 25 m/s, initial velocity is 0 and time is 3 seconds.

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{25-0}{3} \\ \\ :\implies \sf a \: = \dfrac{25}{3} \\ \\ :\implies \sf a \: = 8.33 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 8.33 \: ms^{-2}$

(b) The distance covered in the region ABE.

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: \triangle \\ \\ :\implies \sf Distance \: = Area \: of \: \triangle \: ABE \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times Base \times Height \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times (3-0) \times (25-0) \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 3 \times 25 \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 75 \\ \\ :\implies \sf Distance \: = 37.5 \: m$

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ii) What is the quantity which is measured by the area occupied below the velocity time graph!?

Answer: The quantity which is measured by the area occupied below the velocity time graph is the distance or displacement.

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iii) Write position time equation.

Answer:

${\small{\underline{\boxed{\sf{:\rightarrow \: s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}$