Triangular numbers and v-t graph

[lwymarie]

http://student.shcc.edu.hk/~s021107/phy.GIF

Displacement=area under v-t graph = 150m

using another concept,
the acceleration is 0.75m^s-2, so the displacement at
1s=0.75m
2s=0.75*2m
.
.
.
20s=0.75*20m

total displacement =sum of displacement of every second
=0.75*(1+2+3+...+18+19+20)
=157.5m
Can you tell me wt's wrong here?

 

[Doc Al]

Your link isn't working, but I assume you are describing uniformly accelerated motion starting from rest.

Since $v = at$:

At t = 0s, the speed is 0 m/s
At t = 1s, the speed is 1*.75 = 0.75 m/s
At t = 2s, the speed is 2*.75 = 1.5 m/s, ... etc.

To find the displacement during each second, use $\Delta d = v_{ave} \Delta t$, where the average speed equals $v_{ave} = (v_i + v_f)/2$. Add these displacements and you'll find that they total to 150m.

 

[thepatientmental]

Thank you for sharing the link to the v-t graph and discussing the concept of triangular numbers. The displacement of an object can indeed be calculated by finding the area under the v-t graph, as you mentioned. However, in your calculation of the total displacement, there seems to be a mistake. The formula for the sum of consecutive integers from 1 to n is n(n+1)/2, not n(n-1)/2 as you used. So the correct calculation would be 0.75*(1+2+3+...+18+19+20) = 0.75*210 = 157.5m. This aligns with the given displacement of 150m, showing that the object has not moved in the opposite direction and the v-t graph is consistent with the given information. It's important to check our calculations and formulas to ensure accuracy in our results.