Rectilinear Acceleration

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In today’s class, we will be talking about rectilinear acceleration. Enjoy the class!

Rectilinear Acceleration

Acceleration

If a body changes its velocity with time, it is said to be accelerated. Acceleration is defined as the rate of change of velocity with respect to time. Its unit is m/s².

Suppose a car moves with a particular velocity (u) and then it increases the velocity to (v) in the time interval (t), then the acceleration of the car is given by:

a = v – u / t

if on the other hand, the car slows down from v to u, the acceleration will have a negative value, which indicates deceleration or retardation.

Acceleration is a vector quantity.

Non-uniform acceleration

When the velocity of a moving body increases by an equal amount in equal intervals of time, no matter how small the time interval may be, it is said to move with uniform acceleration.

Acceleration due to gravity

Whenever a body falls freely, its velocity increases steadily from zero. In other words, the body accelerates. Conversely, when a body is thrown upwards, the speed decreases gradually until it comes to a stop (deceleration). This is due to the fact that the earth attracts bodies close to it or on its surface. It has been estimated that the velocity of a body falling freely under gravity increases by about 9.8m/s² every second. In other words, the acceleration due to the gravitational pull of the earth has been estimated to an average of 9.82m/s². This is, however, may vary slightly as one move from one location to another on the surface of the earth.

Equations of uniformly accelerated motion

1^st equation:

By definition, acceleration, a is given by:

a = v – u / t

Where:

v = final velocity

u = initial velocity

t = time taken to change from u to v

we can rearrange the equation above to become

v = u + at

2^nd equation:

For a body whose speed changes from u to v, the average speed is given by:

u + v / 2

Now, distance S = average speed x time

Therefore:

Distance covered, S = ( u + v / 2 )t

If we put v = u + at into the equation above, we have

Distance covered S = ( u + u + at/ 2 )t

S = ( 2u + at / 2 ) t

S =2ut/2 + at²/2

S = ut + ½at² ————————————————————— (2)

3^rd equation:

If we square both sides of equation 1 and expand, we have

v²= (u + at)²

or v²= u² + 2uat + a²t²

v²= u² + 2a (ut + ½at²)

v²= u² + 2aS ————————————————————- (3)

Problem solving with the equations

It is necessary for the student not only to commit the equations of motion to memory but should be able to derive them. To solve problems successfully with the equations, take the following steps:

(a) Identify all the (quantities) in a given problem. That is, you must know what u, v, a, t and s stand for.

(b) You must be able to interpret the given problem to know which of thee quantities (u, v, a, t, s) are given and which one of them is to be calculated.

(d) Ensure all quantities are in S.I units unless otherwise stated. That is for instance if speed is in km/h and time is in second(s), you should convert the unit of speed to m/s. Likewise, if the time given is given in minutes and velocity is given in m/s, the unit of time should be converted to second. Working with inconsistent units will give you the wrong solution to the problem.

(e) Solve the equation for the unknown quantity.

(f) Ensure the solution is expressed in the appropriate (or required) unit.

Worked examples

(1) Find the acceleration of a car, which start from rest and attains a speed of 20m/s in 5 seconds.

Solution

Step 1: Identify the data (quantities) given

u = 0 (velocity is zero for a body at rest)

v = 20m/s

t = 5 Seconds

Step 2: We are to calculate acceleration, a

Step 3: The equation that best connects u, v, t (given quantities) to the unknown quantity is v = u + at

Step 4: All units are consistent (they are in S.I)

Step 5: Solve for a from: v = u + at

20 = 0 + (a x 5)

20 = 5a

A = 4

Step 6: Leaving the answer as in step 5 is meaningless. Acceleration, a, has a unit of m/s². Hence, the answer is given as:

a = 4m/s²

(2) Determine the distance covered by the car, which start from rest and attains a speed of 20m/s in 5 seconds.

Solution

Step 1: Identify the data (quantities) given

u = 0 (velocity is zero for a body at rest)

v = 20m/s

t = 5 Seconds

a = 4m/s²

Step 2: We are to calculate distance, S

Step 3: The equation that best connects u, v, t and a (given quantities) to the unknown quantity is s = ut + ½ at²

Step 4: All units are consistent (they are in S.I)

Step 5: Solve for S from: S = ut + ½ at²

S = 0 x 5 + ½ x 4 x 5²

S = 0 + ½ x 4 x 25

S = 0 + 2 x 25

Step 6: S = 50m

(3) A car travelled from Lagos to Badagry, a distance of 80km. if the journey took 45 minutes, what was the speed of the car? In

(i) Km/h

(ii) m/s

Solution

Distance = 80km

Time = 45 minutes = ¾hr

Speed = distance/time = 80/(3/4)= (80*4)/3 = 106.7 km/hr

To express the answer in m/s, proceed as follows:

Distance = 80km = 80 x 1000m = 80,000m

Time = 45 min = 45 x 60 seconds = 2700 seconds

Speed = m/s

Now, there is a shorter way to convert km/h to m/s and vice – versa:

(i) To convert km/hr to m/s, simply divide the given value by 3.6

e.g. 72km/h = = 20m/s

108km/hr = 30m/s

(ii) Conversely, to convert m/s to km/hr, multiply the given value by 3.6

e.g. 12m/s = 12 x 3.6 km/hr = 43.2 km/hr

5m/s = 5 x 3.6 km/hr = 18km/hr

In our next class, we will be talking about Upthrust (Buoyancy). We hope you enjoyed the class.

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Rectilinear Acceleration