Speed, Velocity and acceleration

Physics SSS2 First Term

Sub Theme: Interaction of matter, space and time

WEEK 3

Speed, Velocity and acceleration

Performance Objectives

Students should be able to;

1. Define speed, velocity and acceleration
2. Show that speed is a scalar quantity while velocity and acceleration are vectors
3. Show on a (V – t) graph the motion of a body
4. Deduce the distance covered between any time intervals on the graph in 3 above

Velocity – Time v-t Graph

Content

1. Gradient of a v-t graph = acceleration

Acceleration = gradient = change in velocitychange in time

acceleration(a) =  

2. Area under a v - t graph = distance.

1. Total distance covered during the motion = area of trapezium 0edc
2. Distance covered during acceleration = area of triangle 0ea
3. Distance covered during constant velocity = area of rectangle aedb
4. Distance covered during deceleration = area of triangle bdc
5. Acceleration = slope of line 0e, a =
6. Deceleration =  slope of dc, −a =

Example 1:

A car starts from rest and accelerates uniformly to 15ms^-1 in 5 s. It then continues at this velocity for the next 10s before decelerating back to rest in another 8 s.

Use the information to answer the following questions

1. Sketch the velocity time graph of the motion of the car
2. Calculate the acceleration of the car
3. Calculate the deceleration of the car
4. What is the total distance travelled by the car
5. Estimate the average speed of the car.

Solution:

1.

2. Acceleration a =

a = 15/5 =3ms⁻²

3. Deceleration –a =

−a = =  

-a = −1.875ms⁻²

4. Total distance = area under the graph = area of trapezium

S = (a + b)h

S =  (15 − 5) + (23 − 0)15

S =  (10 + 23)15

S =  =

S = 247.5m

5. Average speed v = total distancetotal time 

V =

= 10ms⁻¹

Example 2:

A body at rest is given an initial uniform acceleration of 8.0ms² for 30s after which the acceleration is reduced to 5.0ms² for 30s. The body maintains the speed attained for 60s after which it is brought to rest in 20s.

(a) Draw the velocity-time graph of the motion using the information given above.

Using the graph, calculate

(b) Maximum speed attained during the motion. 

(c) Average retardation as the body is brought to rest. 

(d) Total distance travelled during the first 60s 

(e) Average speed during the same intervals as in (c)

Solution

(a)

(b) There are two stages of acceleration

Stage 1: Acceleration = gradient

a = 8ms⁻²

a =

8 =  

Cross multiplying,

V₁ = 8 × 30 = 240ms⁻¹

Stage 2: a = 5ms⁻²

a =

5 =

Cross multiplying,

V₂ − V₁ = 150

But V₁ = 240

V₂ – 240 =150

V₂ = 150 + 240 = 390ms⁻¹

The maximum velocity is 390ms⁻¹

(c) Average retardation is equal to gradient

−a =  

But V₂ = 390ms⁻¹

−a =

−a = 19.5ms⁻²

Average retardation =−19.5ms−2

(d) Distance is in the first 60sec = area of triangle + area of the next trapezium

S =  (time) V₁ +   (V₁+V₂) times

=  (30)(240) +  (240 + 390)30

S = 3600 + 9450

= 13050m

(e) Average speed V =  

V =

= 217.5ms⁻¹

Relative Motion

This is the motion of a body with respect to another. All motion is relative. The motion of a car on the road is with respect to the earth or any other frame of reference in which the motion of the car is being observed.

Resultant Velocity of Relative Motions

• Consider two cars X and Y travelling in the same direction and at the same speed, a commuter in X will observe that Y is stationary (not moving)

V_x = V_y

Relative velocity V_x − V_y = 0

• If car X is to be travelling at a speed V_x which is greater than the speed of V_y, a commuter in car Y will observe the speed of car X to be

V_x − V_y = relative velocity of car X with respect to Y

A commuter in X will observed the relative velocity of Y to be

V_y − V_x

This value will be negative. This means that to an observer in X, the car Y will appear to be going backward (going the opposite direction with a speed of /V_y − V_x/

• But if car X and Y were to be travelling in opposite direction, the relative velocity of X with respect to Y will be

V_x − V_y = relative velocity of X with respect to Y

V_y − V_x = relative velocity of Y with respect to X

N.B. note that the relative velocity of X with respect to Y, V_xy is equal in magnitude but opposite in direction to the relative velocity of Y with respect to X, V_yx.

V_xy = −V_yx

Examples

1. Two racing cars A and B travelling in the same direction at 300m/s and 340mls respectively. What is the relative velocity of A with respect to B?

Solution:

V_a = 300km/h

V_b = 340km/h

Relative velocity of a with respect to B, V_ab = V_a − V_b = 300 – 340

 = −40km/h

(Note that this is negative. A appears to be travelling in the opposite direction to B)

2. A boat whose speed is 8 km/h sets course on a bearing 060⁰. If the tide is running at a speed of 3 km/h from a bearing of 330⁰, find;

3. The actual speed of the boat(i.e, relative speed of the boat)

The direction of travel

To obtain the relative velocity (actual velocity), draw the component velocity such that the head of one point to the end of the other. Draw the relative velocity to beginning from end of the first to the head of the last.

Using Pythagoras theorem

 = 8² + 3²

 = 64 + 9 = 73

 = 73

= 8.54kmh⁻¹

Let ϴ be the angle between the relative velocity and the direction of the boat.

tanθ =  =

θ = tan⁻¹[0.375] = 20.6^o