1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
riadik2000 [5.3K]
2 years ago
11

Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const

ant acceleration Of 1m/s². Q Starts from N at the Same time with velocity 6m/s and at a Constant acceleration of 3m/s². find the time when the
a) Particles are 30m apart, 13
b) Particles meet
c) velocity of P is ¾ of the velocity of Q


​
Physics
1 answer:
bagirrra123 [75]2 years ago
7 0

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

S = ut + \frac{1}{2}at^{2}

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

#SPJ1

You might be interested in
A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0
3 years ago
A steel guitar string with a diameter of 0.300 mm and a length of 70.0 cm is stretched by 0.500 mm while being tuned. How much f
Ganezh [65]

Answer:

10.1 N

Explanation:

Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.

8 0
3 years ago
The standard measure used to compare sound intensities is the ______ .
uranmaximum [27]
The answer is Decibels. <span />
4 0
3 years ago
Read 2 more answers
Genevieve was working on a science lab. In her lab, she was combining two chemicals and timing how long the reaction took to for
dexar [7]

Answer: Her test trials had a high level of accuracy but a low level of precision.

Explanation:

5 0
3 years ago
Ellipses have only one focus. Question 14 options: True False
laila [671]

Answer:

False. This is because ellipses have 2 focus points and not only one.

6 0
2 years ago
Other questions:
  • I need help with 6a, 6b, and 6c !!
    8·1 answer
  • Which of the following is an organic compound?
    7·2 answers
  • Which statement best describes wavelength
    5·2 answers
  • How are hypertension, heart disease, and stroke related?
    14·1 answer
  • A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str
    8·1 answer
  • A harmonic oscillator begins to vibrate with an amplitude of 1.6 m, but after a time of 1.5 minutes, the amplitude has dropped t
    14·1 answer
  • Decide if each description below is an example of a change caused by energy. From the drop-down menu, choose "IS" if it is an ex
    7·1 answer
  • A car travels for 50mph for 5 hours, then for 30mph for another 3 hours in the same direction. What is the average speed the mot
    10·1 answer
  • A car has an acceleration of 4m/s^2 to the east for 5 seconds. What is the change in velocity?
    15·1 answer
  • What is the equation for finding the amount of inertia in forces and motion? Explain how you solve that equation.​
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!