Question

the variation with time t of the acceleration a of an object is shown.
what is the change in velocity of the object from t = 0 to t = 6 s?
a. 6 m s⁻¹
b. 8 m s⁻¹
c. 10 m s⁻¹
d. 14 m s⁻¹

1. a stone is thrown downwards from the edge of a cliff with a speed of 5.0 m s⁻¹. it hits the ground 2.0 s later. what is the height of the cliff?

a. 20 m
b. 30 m
c. 40 m
d. 50 m

Explanation:

Step1: Recall the relationship between acceleration - time graph and velocity - change

The change in velocity $\Delta v$ is equal to the area under the acceleration - time ($a - t$) graph.

Step2: Calculate the area of the trapezoid

The area of a trapezoid is given by $A=\frac{(a + b)h}{2}$, where $a$ and $b$ are the lengths of the parallel sides and $h$ is the height. For the $a - t$ graph from $t = 0$ to $t=6s$, the parallel sides of the trapezoid are $a = 1.0m/s^{2}$ and $b = 3.0m/s^{2}$, and the height $h = 4s$, and there is also a rectangle with base $2s$ and height $1.0m/s^{2}$.
The area of the rectangle $A_{1}=1\times2=2m/s$.
The area of the trapezoid $A_{2}=\frac{(1 + 3)\times4}{2}=8m/s$.
The total area $A = A_{1}+A_{2}=2 + 8=10m/s$.

Answer:

C. $10m/s$