Question

A particle starts from rest and moves with a uniform acceleration of 5 metre per second for 10 seconds and then it moves with a constant velocity of 4 seconds slows down and comes to rest in 5 seconds for the velocity graph for the motion of the objects and answer the following question what is the maximum velocity attained by the body what is the distance travelled during the period of acceleration what is the distance travelled by the body​

Answer 1

Secondary SchoolPhysics 5+3 pts

A body starts from rest and moves with a uniform acceleration of 5m/s2 for 5s and then it moves with a constant velocity for 4s. Later it slows down and comes to rest in 5s. Draw the velocity graph for the motion of the body and answer the following questions:

a) What is the maximum velocity attended by the body?

b) What is the distance travelled during this period of acceleration?

c) What is the distance travelled when the body was moving with constant velocity?

d) What is the negative acceleration of the body while slowing down?

e) What is the distance travelled by negative acceleration?

f) What is the total distance travelled?

Answers

See the diagram enclosed.

At t=0, start at O, origin. Draw a line to A with a slope of 5 for 5 seconds. For every one second move up by 5 units vertically.

Then draw a horizontal line for 4 seconds reaching D. here t= 9sec.

Now join D with E at t= 9+5 = 14sec on time sline when the body stops ie., velocity is 0. The slope here is found by height lost / horizontal span = 25/5 = 5 m/sec²

a) V at A : 25 m/sec

b) Area enclosed by triangle OAB = 1/2 * 25 * 5 = 62.5 m

c) area of ABCD = 25 * 4 = 100 m

d ) slope = (0-25)/5 = -5 m/sec²

e ) area of triangle DCE = 62.5 m

f ) total area bounded between the OADE and time axis. = 62.5 + 100 + 62.5 = 230m

Answer 2

Answer:

The maximum velocity attained by the body= 50 m/s.

Distance travelled during the period of positive acceleration = 250 m.

Distance travelled during the period of positive acceleration = 125 m.

Distance travelled by the body​ = 575 m

Explanation:

• Particle starts from rest. So initial velocity = u = 0

In OA part for time(t) = 10 seconds, the particle's acceleration, a = 5 m/s²

From 1st equation of motion, final velocity v = u + at

                                                                           = 0 + (5 × 10) = 50 m/s

• Now the particle will move with 50 m/s velocity for 4 second. (In graph AB part describes this motion).

        Then the particle slows down. So the maximum velocity = 50 m/s.

• After reaching B point its velocity decreases and within 5 second it comes to rest, which is shown by BC part.
• Distance Calculation:

 Distance travelled during the positive acceleration

                             = area under OA (from graph)

                              = $\frac{1}{2}$ × OD × DA = $\frac{1}{2}$ × 10 × 50 = 250 m.

  Distance travelled during the negative acceleration

                              = area under BC (from graph)

                              = $\frac{1}{2}$ × EC × BE = $\frac{1}{2}$ × 5 × 50 = 125 m.

Distance travelled during constant velocity

                              = area under AB (from graph)

                              =  DE × DA = 4 × 50 = 200 m.

Total distance = 250 + 125 + 200 = 575 m

In attached graph OABC is the motion of the particle.

To get this type of problems visit the link given below:

https://brainly.in/question/1205543

https://brainly.in/question/20707512

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