Describing motion along a line
Describing motion along a line

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Describing motion along a line

6.2 End-of-course questions

Question 25

Table 8 shows the atmospheric pressure P in pascals (Pa) at various heights h above the Earth's surface. Plot a graph to give a visual representation of the data in the table. Be careful to label your axes correctly. Explain why you have chosen to plot particular variables on the horizontal and vertical axes. Use your graph to find the rate of change of atmospheric pressure with height at h = 10 km.

Table 8: Data for question 25

0101 325
548 586
1023 297
1511 171


The graph is given in Figure 43. Notice how the vertical axis is scaled by 105 in order to avoid a confusion of numbers on this axis. Since the pressure is the dependent variable, it is plotted on the vertical axis. Conversely, since height is the independent variable, it is plotted along the horizontal axis.

The rate of change of atmospheric pressure with height at 10 km is given by the gradient of the graph at this point. Using the tangent shown in Figure 43, we obtain

(Differentiating the function that was used to produce Table 8 gives a gradient of −3435 Pa km −1 at 10 km.)

Figure 43: Graph for question 25

Question 26

(a) Define the terms position and displacement, and carefully distinguish between them.

(b) The position x of a particle at time t is given in Table 9. Plot a position-time graph for the data in Table 9.

(c) Using your graph, measure the velocity of the particle at t = 5 s.

(d) A second particle undergoes uniform motion and has the same position and velocity as the first particle at time t = 5 s. What is the displacement of the second particle from the first at time t = 10 s?

Table 9: Data for question 26



(a) The position of a point on a line is represented by a coordinate x, measured from some arbitrarily chosen origin. Such a point might represent the instantaneous position of a particle moving along the line. Displacement refers to the difference in position of two points, these might be the initial and final positions of a moving particle, in which case the displacement would represent the change in the particle's position.

(b) The position-time graph for Table 9 is given by the curve in Figure 44.

(c) The instantaneous velocity is given by the gradient of the tangent at t = 5 s. The gradient of the tangent shown is

(This result compares with the value of 1.087, which is obtained by differentiating the function that was used to produce Table 9.)

(d) The displacement-time graph for the second particle is the tangent at t = 5 s, which we have already drawn in Figure 44 since for uniform motion x = x0 + vxt. From this tangent, we can see that the position of the second particle at t = 10 s is 8.5 km. Hence, the displacement of the second particle from the first is (8.5 − 12.8) km = −4.3 km. Note that the minus sign is an essential part of the answer.

Figure 44: Position-time graph for question 26

Question 27

(a) Define the terms velocity and acceleration.

(b) The velocity vx of a particle moving along the x-axis at various times t is given in Table 10.

(i) Assuming the particle has a constant acceleration between the given positions, draw a velocity-time graph representing the data in Table 10.

(ii) Use your graph to calculate the total displacement of the particle over the time interval given in the table.

Table 10: Data for Self-assessment question 27

v x /m s−1t/s


(a) The velocity of a particle is its rate of change of position. The acceleration of a particle is its rate of change of velocity.

(b) (i) The velocity-time graph for Table 10 is given in Figure 45.

(ii) The displacement is equal to the signed area under the velocity-time graph. For the interval between 0 s and 10 s, the areas above and below the time axis cancel and consequently the displacement is zero.

For the interval between 10 s and 15 s, the displacement is given by

−(1/2) x 5 x 4 m = −10 m.

For the interval between 15 s and 20 s, the displacement is given by

(1/2) x 5 x 2 m = 5 m.

The total displacement is the sum of all three contributions:

(0 − 10 + 5) m = −5 m.

Figure 45: Velocity-time graph for question 27

Comment In questions 28 to 32, you are not required to draw any graphs.

Question 28

The variable z is related to the variable y by the equation

z = 3 + 2y + y3.

(a) Find the derivative dz/dy.

(b) Evaluate dz/dy at y = 2.

(c) What would be the gradient of a graph of z plotted against y for the value y = 2?


(a) dz/dy = 2 + 3y2.

(b) Substituting y = 2 into the derivative found in part (a) we obtain


(c) Since the gradient is equal to dz/dy evaluated at y = 2, the answer is 14.

Question 29

A rocket travels vertically away from the surface of the Moon. It is still close to the Moon's surface when it jettisons an empty fuel tank. The fuel tank initially travels with the same velocity as the rocket, but falls back to the Moon, reaching the Moon's surface 50 s after being released. If the fuel tank hits the surface at a speed of 50 m s−1, calculate the speed of the fuel tank when it was jettisoned. You may assume that the magnitude of the acceleration due to gravity near the Moon's surface is 1.6 m s−2.


Choose an x-axis pointing vertically upwards from the Moon's surface. Using Equation 28b

with vx = −50 m s−1, ax = −1.6 m s−2 and t = 50 s, we obtain

ux = vxaxt = (−50 + 1.6 × 50) m s−1 = 30 m s−1.

Question 30

For time t greater than or equal to zero, a particle's position as it travels along the x-axis is described by the function x(t) = At 2, where A = 4.0 m s−2. Use differentiation to calculate how fast the particle is travelling at t = 10 s.


and therefore

vx (10 s) = (2 × 4.0 × 10) m s−1 = 80 m s−1.

Question 31

A vase falls to the ground from a shelf at height 1.80 m above the floor. Neglecting air resistance, calculate the time taken for the vase to strike the floor.


Taking up as the positive direction, and using Equation 28a

with ux = 0 m s−1, ax = −9.81 m s−2 and sx = −1.80m, we obtain

Question 32

(This question is more difficult than its predecessors.) A rocket is initially at rest on the Earth's surface. At time t = 0 s the rocket motor is fired and the rocket accelerates vertically upwards at a constant 2.00 m s−2. After a time interval of 20.0 s the motor fails completely and the rocket accelerates back to Earth with a downward acceleration of magnitude 9.81 m s−2.

(a) Calculate the height reached by the rocket at the instant the motor fails.

(b) Calculate the velocity of the rocket at the instant the motor fails.

(c) Find an equation that, for situations involving constant acceleration, gives the displacement in terms of the initial velocity, the final velocity, and the constant acceleration.

(d) Calculate the distance travelled by the rocket from the instant the motor fails until the rocket reaches its highest point. Hence, find the total height gained by the rocket. [Hint: The equation you obtained in part (c) may be useful here.]

(e) Find the total time taken for the rocket to fall back to Earth from its highest point.


(a)Taking up as positive, and using Equation 28a

with ux = 0 m s−1, ax = 2.00 m s−2 and t = 20.0 s, we obtain

sx = (1/2) × 2 × (20.0)2 m = 400 m.

This is the displacement from the initial position when the motor fails. It follows that the height at which the motor fails is s = |sx | = 400 m.

(b) Using Equation 28b

vx = ux + axt

with ux = 0 m s−1, ax = 2.00 m s−2 and t = 20.0 s, we obtain

vx = 0 + 2 × 20.0 m s−1 = 40.0 m s−1.

(c) As shown in Section 5.2, eliminating t from Equations 28a and 28b gives Equation 28c

which we can write as

(d) Substituting vx = 0 m s−1, ux = 40.0 m s−1 and ax = −9.81 m s−2 in the above equation, we obtain

Adding this result to the displacement obtained in part (a) we obtain a total height of 482 m.

(e) Using Equation 28a

with ux = 0 m s−1, ax = −9.81 m s−2 and sx = −482 m, we obtain


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