The velocities, v ms^{-1}, of an accelerating car are recorded as a function to time, t s, in the following table.

t /s

3.0

4.0

5.0

6.0

7.0

v /ms^{-1}

11.4

14.8

18.0

21.0

23.9

Given that acceleration is the rate of change of velocity, estimate the acceleration of the car at time, t = 5 s.

3.60 ms^{-2}

3.13 ms^{-2}

3.10 ms^{-2}

3.00 ms^{-2}

The solid line in the graph shows the velocity, v ms^{-1}, of a car with time, t s.

Using the tangent drawn at time, t = 5 s, estimate the acceleration at t = 5 s, given that acceleration is the rate of change of velocity.

2.20 ms^{-2}

1.96 ms^{-2}

1.78 ms^{-2}

1.68 ms^{-2}

If y = 2x^{3}, which of the following equations gives the rate of change of y as a function of x?

dy/dx = 5x^{2}

dy/dx = 6x^{2}

dy/dx = 6x^{4}

dy/dx = 3x^{2}

The voltage, V v, on a discharging capacitor at a time, t s, is given by V = 5.6 e ^{-1.4t}. Select ALL of the following statements that are correct: (Click here to view the Video Answer)

If v = 3×√s, which of the following equations gives the rate of change of v as a function of s?