Everything Maths Grade 12

CHAPTER 7. DIFFERENTIAL CALCULUS 7.6

Velocity after 1 ,5 s:

v(1,5) = 20− 10(1,5)

= 5 m· s−^1

Step 3 : Zero velocity

v(t) = 0

20 − 10 t = 0

10 t = 20

t = 2

Therefore the velocity iszero after 2 s

Step 4 : Ground velocity

The ball hits the groundwhen h(t) = 0

20 t− 5 t^2 = 0

5 t(4−t) = 0

t = 0 or t = 4

The ball hits the groundafter 4 s. The velocity after 4 s will be:

v(4) = h�(4)

= 20− 10(4)

=−20 m· s−^1

The ball hits the ground at a speed of 20 m· s−^1. Notice that the sign of the

velocity is negative which means that the ball is moving downward (the reverse

of upward, which is when the velocity is positive).

Step 5 : Acceleration

a = v�(t)

=−10 m· s−^1

Just because gravity is constant does not meanwe should necessarily think of

acceleration as a constant. We should still consider it a function.

Chapter 7 End of Chapter Exercises

1. Determine f�(x) from first principles if: