Motion in a Plane with constant Acceleration

Motion in a Plane with Constant Acceleration

The study of motion is called mechanics. The study of motion has two parts: kinematics and dynamics. From kinematics, the equation of motion is obtained. A body under uniform acceleration is considered here. A body has two kinds of acceleration. One is the change in velocity, while the other is a change in direction. In this case, the acceleration is uniform and constant. This blog is about Motion in a Plane with constant acceleration wherein we will study Motion in a Plane, acceleration, constant acceleration, formula, and some sample problems. 

Motion in a Plane

Motion in a Plane is the Motion in two dimensions that requires a two-coordinate system, such as an X-Y coordinate system. An instance of such movement is in the path of a projectile or circular motion.

A common example of motion in a plane is that of projectile motion. In terms of physics, an object in flight is said to be in uniform (unaccelerated) motion if the speed it’s traveling at is unchanging over time.

Acceleration

Acceleration is the rate of change of an object’s velocity. Acceleration can be either constant or varying over time and in the case of constant acceleration, its value is a combination of the net change in velocity and total time over which this occurred. It is also called average acceleration.

Constant acceleration

Constant acceleration is a type of acceleration in which the rate of change of velocity remains constant with time. In this case, if one were to graph the movement of an object, it would appear as a straight line. Since acceleration is a vector quantity and motion is always in a single direction, it does not matter whether speed or distance are used as the identification for a vector. 

Also, since all vectors have magnitude (how much they move), it can be said that with constant acceleration, physical change isn’t gradually changing directions but rather “speeding up or slowing down while keeping in the same direction indefinitely.” 

Constant acceleration formula or equations

Keeping the value of acceleration constant, the equation of motion can be described. Let’s say the original haste of an object was “ u”, now a constant force is applied which causes the body to move with constant acceleration “ a” and the body reaches the haste in time “ t” while covering the distance “ s”. 

First Equation of Motion

In the case of constant acceleration, its value is given by, 

v = u + at

Second Equation of Motion 

Instantaneous velocity is given by, 

v =ds/dt

This equation can be rearranged in the following form,

ds = v*dt

Substituting the value of velocity we get, 

ds = (u+at)dt

Integrating both sides we will get, 

s = ut + 1/ 2 at2

Third Equation of Motion

Instantaneous acceleration and instantaneous velocity is given by, 

a = dv/dt

v = ds/dt

Cross multiplying both of these equations we will get, 

adsdt = v dvdt

a ds = v dv

After integration we will get

a s = v2 – u22

v2 = u2 + 2as

Note: These equations can only be used when the acceleration is constant

Sample Problems

1. A car moves on a  straight road  at 20 m/s, and then increases its speed from it’s starting speed  to 50 m/s with acceleration of 1.5 m/s2, how much time it needs to reach at speed 30m/s and how much distance it will travel in this time? 

Solution:

Firstly consider what information has been given, 

u = 20 m/s

v = 50 m/s 

a = 1.5 m/s2

The question asks for the values of time and then speed.

We know that:

v = u + at

Inserting the known values into the equation gives:

50 = 20 + 1.5t

30 = 1.5t

⇒ t = 20 s

To calculated s we have,

s = 1/2 (u + v)t

s = 1/2 (20 + 50) × 20

s  = 700 m

2. A boy throws a ball vertically upwards from the ground at 5.5 m/s speed. Calculate the maximum height reached by the ball.Yo You can neglect air resistance and consider g = 9.8 m/s2 .

Solution:

It is known that 

u = 5.5 m/s 

a = -g = −9.8 m/s2 

as the gravity acts in a downwards direction and the positive direction is upwards. It can be understood that at the maximum height v = 0 m/s

Therefore, using:

v2 = u2 + 2as

0 = 5.52 + 2 × (−9.8) × s

0 = 30.25 − 19.6 × s

⇒ s = 1.54 m

Conclusion

We hope you enjoyed our article about Motion in a Plane with Constant Acceleration. This blog was about Motion in a Plane with constant acceleration which covered  Motion in a Plane, acceleration, constant acceleration, its graph, formula, and some sample problems related to it. Hope you may have liked this blog.Thanks for reading! We are always excited when one of our posts can provide useful information on topics like this.