Question

QIL A Cheetah starts from rest and accelerates at 2m/s for 10 seconds. Calculate
1 Final Velocity
2. Distance travelled
Q2. A train traveling at 20 m/ accelerate at 0.5m/ for 30 seconds. How far will it
travel in this time
03. A cyclist is travelling at 15m/s. She applies brakes so that she does not collide
with a wallem away What deceleration must she have?
04. Starting from rest, Aman ride his scooter to attain a velocity of 10 m/s in 40s.
Then he applies brales such that velocity comes down to 6m/s in Calculate
socderation of scooter in both the cases.
05. Practice all the graphs in NCERT boble from chapter motion. Draw and Paste in
holiday homework notebooles
06. Derive using graphical method vrucat
s-ut lat
2

Answer 1

Answer 1st:

ProvidEd that:

• Initial velocity = 0 m/s
• Acceleration = 2 m/s²
• Time taken = 10 seconds
• Final velocity = ???
• Distance travelled = ???

SolutioN:

• Final velocity = 20 m/s
• Distance travelled = 100 m

RequirEd solution:

~ Firstly finding final velocity, v by using first equation of motion!

⇒ v = u + at

⇒ v = 0 + 2(10)

⇒ v = 0 + 20

⇒ v = 20 m/s

~ Now let's calculate the distance, s by using third equation of motion!

⇒ 2as = v² - u²

⇒ 2(2)(s) = (20)² - (0)²

⇒ 4s = 400 - 0

⇒ 4s = 400

⇒ s = 400/4

⇒ s = 100 m

____________________________

Answer 2nd:

ProvidEd that:

• Acceleration = 0.5 m/s²
• Initial velocity = 20 m/s
• Time taken = 30 seconds
• Distance = ???

SolutioN:

• Distance = 825 m

RequirEd solution:

~ Let's find the distance, s by using second equation of motion!

⇒ s = ut + ½ at²

⇒ s = 20(30) + ½ × 0.5(30)²

⇒ s = 20(30) + ½ × 0.5 × 30 × 30

⇒ s = 20(30) + ½ × 0.5 × 900

⇒ s = 20 × 30 + ½ × 0.5 × 900

⇒ s = 600 + ½ × 0.5 × 900

⇒ s = 600 + ½ × 450

⇒ s = 600 + 225

⇒ s = 825 m

____________________________

Answer 3rd

ProvidEd that:

• Initial velocity = 15 m/s
• Final velocity = 0 m/s

To calculAte:

• The deceleration

Please read here: I can't solve this question because the time isn't given here as time is required to solve this question. Thanks for understanding!

____________________________

Answer 4th

ProvidEd that:

• Final velocity = 10 m/s
• Initial velocity = 0 m/s
• Time taken = 40 seconds

Then Aman applied a break that velocity comes down to 6 m/s in same time

To calculAte:

• The acceleration in both cases

Required solution:

#InTheCase1st

⇒ a = (v-u)/t

⇒ a = (10-0)/40

⇒ a = 10/40

⇒ a = 1/4

⇒ a = 0.25 m/s sq.

#InTheCase2nd

⇒ a = (v-u)/t

⇒ a = (6-10)/40

⇒ a = -4/40

⇒ a = -1/10

⇒ a = -0.1 m/s sq.

Please read here: In this question I find the acceleration in proper way but when finding acceleration in second case, dear user you haven't mentioned the time when break applied so I take the same time but according to that answer is correct. Thanks for understanding!

____________________________

Answer 5th:

Please read here: I can't do this question as here you have to practice the graphs by your own to make your concepts better!

____________________________

Answer 6th

We are asked to derive second Equation of motion that is s = ut + ½ at² Let's derive!

~ Firstly according to the graph,

⇢ AO = DC = u (Initial velocity)

⇢ AD = OC = t (Time)

⇢ EO = BC = v (Final velocity)

~ Now let, the object is travelling a distance in time under uniform acceleration. Now according to the graph we are able to see that inside the graph the obtained area enclose within OABC(trapezium) under velocity time graph AB. Therefore, distance travelled by an object can be given by

$\tt \Rightarrow Distance \: = Area \: enclosed \\ \\ \tt \Rightarrow s \: = Area \: enclosed \\ \\ \tt \Rightarrow s \: = OABC \\ \\ \tt \Rightarrow s \: = Area \: of \: rectangle \: + Area \: of \: \triangle \\ \\ \tt \Rightarrow s \: = Length \times Breadth + \dfrac{1}{2} \times Base \times Height \\ \\ \tt \Rightarrow s \: = AO \times AD + \dfrac{1}{2} \times AD \times BD \\ \\ \tt \Rightarrow s \: = u \times t + \dfrac{1}{2} \times t \times BD \\ \\ \tt \Rightarrow s \: = u \times t + \dfrac{1}{2} \times t \times at \\ \\ \tt \Rightarrow s \: = ut + \dfrac{1}{2} \times at^2 \\ \\ {\pmb{\sf{Henceforth, \: derived!}}}$

How the value of BD came?

Firstly we can write BC as BD + DC. Now as BD have the velocity position and DC have the time position. Henceforth, we already know that

$\tt \Rightarrow Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{v-u}{t} \\ \\ \tt \Rightarrow a \: = \dfrac{BD}{t} \\ \\ \tt \Rightarrow at \: = BD$