Question

(b) The speed vs. time graph below shows information the journey of a toy car during a period of 80 seconds. Speed(m/s) 120 ring В C 100 80 60 20 A D Time(s) 10 20 50 40 50 60 7 80 90 Calculate the acceleration of the toy car between point A to point B. [2] (ii) What was the total distance covered by the toy car? [2] (iii) What was the average speed of the toy car? [1]

Answer 1

Answers:

1)

`a_(AB)=5(m)/(s^2)`

2)

`d=6000m`

3)

`v_{\text{avg}}=75(m)/(s)`

Step-by-step explanation:

1)

The speed of the car at time t=20s is 100m/s (point B) and the speed of the car at time t=0s is 0m/s (point A).

The acceleration of a particle is equal to the change in its velocity divided by the time that it takes for the particle to change its velocity:

`a=(\Delta v)/(\Delta t)=(v_f-v_0)/(t_f-t_0)`

In this case, the acceleration between points A and B on the graph is:

`a_(AB)=(v_B-v_A)/(t_B-t_A)=(100(m)/(s)-0(m)/(s))/(20s-0s)=(100(m)/(s))/(20s)=5(m)/(s^2)`

2)

The distance traveled by a particle on a given time interval is equal to the area under the speed vs time graph. Notice that ABCD is a trapezoid, the area of a trapezoid is equal to the sum of its smaller base and its larger base, multiplied by the height of the trapezoid and divided by two:

`A=((B+b)h)/(2)`

In this case, AD is the larger base, BC is the smaller base and the height is 100m/s. BC=40s and AD=80s. Then, the distance traveled by the car is:

`d=((40s+80s)\cdot100(m)/(s))/(2)=6000m`

3)

The average speed of a particle is equal to the total distance traveled by the particle divided by the time that it takes for it to travel that distance. The car traveled 6000m in 80s. Then, the average speed of the car is:

`v=(6000m)/(80s)=75(m)/(s)`