Question

A car in stop-and-go traffic starts at rest, moves forward 21 m in 8.0 s, then comes to rest again The velocity-versus-time plot for this car is given in the figure (Figure 1). Part A What distance does the car cover in the first 4.0 seconds of its motion? Part B What distance does the car cover in the last 2.0 seconds of its motion? Part C What is the constant speed V that characterizes the middle portion of its motion?

Answer 1

we know,

the area under the velocity time graph shows the distance moved,

the total distance travelled = 21m = area under the graph

area under the graph = area of the trapezium = 1/2 * (sum of parallel side) * height

= 1/2 (2+8) * V = 5V

hence 5V = 21

hence V= 4.2 m/s

for PART A

the distance cover = area of first triangle = 1/2 * 4 * 4.2 =8.4 m

PART B

the distance cover = area of second triangle = 1/2 * 2 * 4.2 =4.2 m

PART C

already calculated V = 4.2 m/s

Answer 2

a)3

b)6

c)2