DISTANCE, DISPLACEMENT AND VELOCITY-TIME GRAPH

 

DISTANCE, DISPLACEMENT, VELOCITY- TIME GRAPH

 DISPLACEMENT: is defined as the distance travelled in a specified direction. It is a vector quantity.

SPEED: is a distance travelled in unit time. It is a scalar. If an object covers equal distances in equal intervals of time. If an object covers equal distances in equal intervals of time, then the object is said to have a constant or uniform speed.

VELOCITY: is the rate of change of displacement with time. It is a vector quantity.

ACCELERATION: is the rate of change of the velocity with time.

VELOCITY-TIME GRAPH

 

 

 

 

 

 

 

 

 

This is the velocity-time graph which depicts the motion of the object during this interval of time. Along OA the body is said to be moving with uniform acceleration because its velocity is increasing by equal amounts in equal intervals of time.

Acceleration along OA =  =  = gradient of the line OA.

Along AB the body maintains the same velocity over the time interval from 5-7 seconds. The body is said to be travelling with uniform or constant velocity. The acceleration during this period is zero. The gradient of the line AB is zero.

Uniform acceleration is defined as the motion of an object whose velocity increases by equal amounts in equal intervals of time, however small the time interval may be eg, motion of body falling under gravity.

Uniform or constant velocity is defined as the velocity of a moving body where equal distances are covered in equal time intervals however small the interval may be.

Along BC, the velocity decreases uniformly from 10 to 0 over the period from 7 seconds to 12 seconds. The body is said to experience retardation or deceleration along this line.

Uniform retardation means that the velocity decreases by equal amounts in equal times, no matter how small the times may be.

Deceleration along BC =  =  = gradient of the line BC

Retardation or deceleration is regarded as negative acceleration.

Variable acceleration: This is when a graph shows an irregular velocity variation.

To calculate the distance from velocity-time graph

Total distance covered is given by the area under the velocity-time graph or the area between the velocity-time graph and the time axis.

Other types of velocity-time graphs

i.                      A ball thrown vertically upwards

The velocity decreases uniformly from OP to zero as the ball moves up its highest level at time A and then increases uniformly as the ball returns to the ground level at time B. Thus the ball first undergoes deceleration or retardation and then an acceleration.

 

 

 

 

 

 

ii.                     A lift moving from one floor to another floor of a tall building

The velocity of the lift increases to a maximum value and then decreases to zero as the lift moves from rest at one floor to a stop at the other floor.   

 

 

 

 

 

 

Equations of Uniformly  Accelerated Motion.

The equations is derived from a body travelling along a straight line with uniform acceleration.

1st  equation of motion: v = u + at

                Acceleration (a)  =  =  

                V = u +at

2nd  equation of motion

Let s be the distance covered in a time t, with the initial and final velocities given by u and v respectively.

Average velocity v =      , from equation one , v =u +at, hence

v =     = u +    at

Distance (s) covered = average velocity x time;  s = (= u +    at)t

S =  ut +    at2

3rd Equation of motion

To obtain this equation both 1st and 2nd equations are used.

From the 1st equation, v = u +at ; 2nd equation, S =  ut +    at2

 square the equation

 V2 = u2 + a2t2

V2 = u2 +2a (ut +    at2 )

V2 = u2 +2as

Example

1.        A body which is uniformly retarded comes to rest in 10s after travelling a distance of 20m. calculate its initial velocity.

Solution

 

 

 

 

S = ½ bxh;  20m = ½ x 10s x u;  40m = 10s x u;   u = 40m / 10s ; u = 4ms-1.

2.        A mango fruit drops to the ground from the top of its tree which is 5m high.  How long does it take to reach the ground?.

Solution

U =0

h = ut +    gt2

5m = 0 +    x10t2 ;   10 = 10t2; t2 =10/10; t=1.0s

3.        A train starts from rest from a station and travels with uniform acceleration 0.5ms-2 for 20s. it travels with uniform velocity for another 30s, the brakes are then applied so that a uniform retardation is obtained and the train comes to rest in a further 10s. Sketch the velocity-time graph of this motion. Using your graph calculate the total distance travelled by the train.

Solution

V= u + at;  v = 0 + (0.5ms-2 x 20s); v= 10ms-1

 

 

 

 

 

S= ½ (a + b)h;   s = ½ (30+60)10; s = 5x90 ; s = 450m.

Assignment

A body at rest is given an initial uniform acceleration of 8.0ms-2 for 30s after which the acceleration is reduced to 5.0 ms-2 for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-time graph of the motion using the information given above.

b. using the graph calculate the:

i. maximum speed attained during the motion.

ii. average retardation as the body is being  brought  to  rest.

iii. total distance travelled during the first 50s.

iv. average speed during the same interval as in (iii)

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