Question

answer this question.. ​

Answer 1

According to the velocity-time graph we are asked to calculate the acceleration and distance covered in 15 seconds, also we have to identify the type of motion.

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!

Required solution:

a) Type of motion: The given velocity time graph shows the uniform motion

b) Acceleration:

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20-20}{25-0} \\ \\ :\implies \sf a \: = \dfrac{0}{25} \\ \\ :\implies \sf a \: = 0 \: ms^{-2} \\ \\ :\implies \sf a \: = 0 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 0 \: ms^{-2}$

Explanation: If the slope of Velocity Time graph is parallel to time axis then the acceleration is always zero as there is no change in velocity as acceleration is the term that tell us about the change in velocity with respect to time.

c) Distance covered in 15 seconds

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: of \: rectangle \\ \\ :\implies \sf s \: = Length \times Breadth \\ \\ :\implies \sf s \: = (15-0) \times (20-0) \\ \\ :\implies \sf s \: = 15 \times 20 \\ \\ :\implies \sf s \: = 300 \: m \\ \\ :\implies \sf Distance \: = 300 \: m$