Question

an object is thrown upwards. the graph shows the variation with time t of the velocity v of the object. what is the total displacement at a time of 1.5 s, measured from the point of release? a 0 m b 1.25 m c 2.50 m d 3.75 m

Explanation:

Step1: Recall the displacement - velocity - time relation

The displacement $s$ is given by the area under the velocity - time graph.

Step2: Divide the graph into geometric shapes

We can divide the $v - t$ graph from $t = 0$ to $t=1.5\ s$ into two triangles.
The first triangle from $t = 0$ to $t = 0.5\ s$ has base $b_1=0.5\ s$ and height $h_1 = 5\ m/s$. The area of a triangle is $A=\frac{1}{2}bh$. So, $A_1=\frac{1}{2}\times0.5\times5 = 1.25\ m$.
The second triangle from $t = 0.5\ s$ to $t = 1.0\ s$ has base $b_2 = 0.5\ s$ and height $h_2=- 5\ m/s$. Its area $A_2=\frac{1}{2}\times0.5\times(-5)=-1.25\ m$.
The third triangle from $t = 1.0\ s$ to $t = 1.5\ s$ has base $b_3=0.5\ s$ and height $h_3 = 5\ m/s$. Its area $A_3=\frac{1}{2}\times0.5\times5 = 1.25\ m$.

Step3: Calculate the total displacement

The total displacement $s=A_1 + A_2+A_3$.
\[s=1.25-1.25 + 1.25=1.25\ m\]

Answer:

B. $1.25\ m$