Graphs in Practical Situations — Solutions

Full worked solutions, with method and final answer for every part.

Question 1

(a)Gradient = change in distance ÷ change in time.

100 ÷ 2 = 50; the gradient represents the speed, 50 km/h

Question 2

(a)Gradient = change in speed ÷ change in time.

20 ÷ 5 = 4; the gradient represents the acceleration, 4 m/s²

Question 3

(a)Speed = gradient = distance ÷ time.

100 ÷ 4 = 25 km/h

Question 4

(a)Gradient = change in speed ÷ change in time.

(10−40)/10 = −3 m/s²; represents the deceleration (3 m/s²)

Question 5

(a)A horizontal section means distance is not changing over time — the object is stationary.

The object is stationary (not moving) during that time

Question 6

(a)Distance travelled equals the area enclosed between the line and the time-axis.

Calculate the area under the speed-time graph

Question 7

(a)A horizontal line has a gradient of zero.

Gradient = 0; represents the object being stationary

Question 8

(a)Substitute x=−1.

y = 3(−1) − 2 = −5

Question 9

(a)Area = speed × time (a rectangle).

20 × 5 = 100 m; this represents the distance travelled

Question 10

(a)A negative gradient would mean distance from the start point is decreasing, i.e. the object is moving back toward its starting position.

The object is moving back toward its starting point

Book a free trial class

Tell us a little about your requirement.