Page 44 - Physics Coursebook 2015 (A level)

Cambridge International AS Level Physics
Summary
■■ Acceleration is equal to the rate of change of velocity.
■■ Acceleration is a vector quantity.
■■ The gradient of a velocity–time graph is equal to acceleration:
∆v a = ∆t
■■ The area under a velocity–time graph is equal to displacement (or distance travelled).
■■ The equations of motion (for constant acceleration in a straight line) are:
■■ Vectors such as forces can be resolved into components. Components at right angles to one another can be treated independently of one another. For a velocity v at an angle θ to the x-direction, the components are:
x-direction: v cos θ
y-direction: v sin θ
■■ For projectiles, the horizontal and vertical components of velocity can be treated independently. In the absence of air resistance, the horizontal component
of velocity is constant while the vertical component of velocity downwards increases at a rate of 9.81 m s−2.
1 v=u+at s=ut+2at
2
s = (u + v)t v2 = u2 + 2as 2
End-of-chapter questions
1 A motorway designer can assume that cars approaching a motorway enter a slip road with a velocity
of 10 m s−1 and reach a velocity of 30 m s−1 before joining the motorway. Calculate the minimum
length for the slip road, assuming that vehicles have an acceleration of 4.0 m s−2. [4]
2 A train is travelling at 50 m s−1 when the driver applies the brakes and gives the train a constant deceleration
of magnitude 0.50 m s−2 for 100 s. Describe what happens to the train. Calculate the distance travelled
by the train in 100 s. [7]
3 A boy stands on a cliff edge and throws a stone vertically upwards at time t = 0. The stone leaves his hand at 20 m s−1. Take the acceleration of the ball as 9.81 m s−2.
a Show that the equation for the displacement of the ball is:
s=20t−4.9t2 [2]
b What is the height of the stone 2.0 s after release and 6.0 s after release? [3]
c When does the stone return to the level of the boy’s hand? Assume the boy’s hand does not move
vertically after the ball is released. [4]
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