Everything Science Grade 10

CHAPTER 21. MOTION IN ONE DIMENSION 21.6

For 0 – 5 seconds: The displace-

ment is equal to the area of the

triangle on the left:

Area△ =^12 b×h

=^12 × 5 s× 4 m·s−^1

= 10m

For 5 – 12 seconds: The dis-

placement is equal to the area of

the rectangle:

Area = ℓ×b

= 7s× 4 m·s−^1

= 28m^2

For 12 – 14 seconds the displace-

ment is equal to the area of the

triangle above the time axis on

the right:

Area△ =^12 b×h

=^12 × 2 s× 4 m·s−^1

= 4m

For 14 – 15 seconds the displace-

ment is equal to the area of the

triangle below the time axis:

Area△ =^12 b×h

=^12 × 1 s× 2 m·s−^1

= 1m

Step 3:Determine the total distance of the car

Now the total distance of the car is the sum of all of these areas:

D = 10m+ 28m+ 4m+ 1m

= 43m

Step 4:Determine the total displacement of the car

Now the total displacement of the car is just the sum of all of these

areas. HOWEVER, because in the last second (fromt= 14s tot= 15

s) the velocity of the car is negative, it means that the car was going

in the opposite direction, i.e. back where it came from! So, to find the

total displacement, we have to add the first 3 areas (those with positive

displacements) and subtract the last one (because it is a displacement in

the opposite direction).

∆~x = 10m+ 28m+ 4m− 1 m

= 41m in the positive direction

Physics: Mechanics 421