Question

please ans this question​

Answer 1

Given:-

• A velocity time graph of a body

To find:-

• Type of motion represented by OA.
• Type of motion represented by AB.
• Type of motion represented by BC.
• Acceleration of the body.
• Retardation of the body

Answer:-

$\maltese \underline{\underline{\sf{1st\:answer:}}}$

• From the graph we can observe that, the slope of the graph is positive and constant. So the rate of change of velocity is positive and constant. Hence we can conclude that the motion is uniformly positively accelerated motion.

$\maltese \underline{\underline{\sf{2nd\:answer:}}}$

• From the graph we can observe that, the slope of the graph is 0. So there is no change in velocity. Hence we can conclude that the body is undergoing constant velocity.

$\maltese \underline{\underline{\sf{3rd\:answer:}}}$

• From the graph we can observe that, the slope of the graph is negative and constant. So the rate of change of velocity is negative and constant. So we can conclude that the motion is uniformly retarded motion/uniformly negatively accelerated motion.

$\maltese \underline{\underline{\sf{4th\:answer:}}}$

• We have concluded earlier that the body is positively accelerated only in the part OA.
• We know that, acceleration = slope = tan∅ = (y₂ - y₁)/(x₂ - x₁)

Here,

• y₂ = 6
• y₁ = 0
• x₂ = 4
• x₁ = 0

So,

Acceleration = (y₂ - y₁)/(x₂ - x₁)

→ Acceleration = (6 - 0)/(4 - 0)

→ Acceleration = 6/4 m/s²

→ Acceleration = 1.5 m/s²

$\maltese \underline{\underline{\sf{5th\:answer:}}}$

• We have concluded earlier that the motion is retarded/negatively accelerated only in part BC.
• We know that, acceleration = slope = tan∅ = (y₂ - y₁)/(x₂ - x₁)

Here,

• y₂ = 0
• y₁ = 6
• x₂ = 14 (considering time interval of OA = BC)
• x₁ = 10

So,

Acceleration = (y₂ - y₁)/(x₂ - x₁)

→ Acceleration = (0 - 6)/(14 - 10)

→ Acceleration = -6/4 m/s²

→ Acceleration = -1.5 m/s²

But if acceleration = -a m/s², then retardation = a m/s²

→ Retardation = 1.5 m/s²