Read displacement off a velocity-time graph
Scenario
Aapke paas ek car ka speed record hai (kitne time pe kitni speed thi). Par ab aapko ye nikalna hai ki gadi ruki kahan, yani total safar kitna lamba tha.
Velocity aur time ke graph se total cover kiya hua raasta (displacement) kaise dhoondh sakte hain?
Mental model
Graph ki line aur x-axis ke beech ka ghera hua 'Area' hi actual me chala gaya distance hai.
Formula `s = v * t` hota hai. Graph me `v` aapki vertical height (`y`) hai aur `t` horizontal lambaai (`x`). Jab aap inko multiply karte ho (jaise rectangle ka area `l * b`), to result me Area aata hai. Yahi area direct gadi ki displacement hota hai.
Explanation
Velocity-time (`v-t`) graph kinematics ka sabse versatile (bahu-upyogi) tool hai. Iske slope se humein acceleration mil jata hai, par us se bhi bada raaz is line ke theek neeche chupa hota hai. Kisi bhi time interval mein, `v-t` graph ki line aur horizontal time-axis (`x`-axis) ke beech jo aakaar (shape) banta hai, us shape ka Area calculate karne par humein us object dwara cover ki gayi displacement mil jati hai.
Ise samajhne ke liye uniform velocity ka case lein: ek car `20 m/s` se `5 s` tak chal rahi hai. Graph par ye ek horizontal line banayegi `y = 20` par. Iske theek neeche `x`-axis tak jo shape banega wo ek rectangle (`20` height, `5` length) hoga. Us rectangle ka area nikalne par (`20 * 5`) humein `100` milta hai. Reality me bhi distance nikalne ke liye `s = v * t` lagane par `100 m` hi aata hai. Isliye 'Area under `v-t` graph = Displacement'.
Agar motion uniformly accelerated hai (starting from rest), to `v-t` graph ek teerchi line banayega origin se. Iske neeche jo shape banega wo ek Right-angled Triangle hoga. Is case me aap displacement nikalne ke liye Triangle ka area formula lagayenge: `1/2 * base * height`.
Kayi baar real scenarios me shape thodi ajeeb (Trapezium) hoti hai. Aise me shape ko tod lena better hota hai. Ek rectangle aur uske upar rakha ek triangle. Dono ka area alag alag calculate karein aur aapas me add kar dein. Yehi trick Class 9 me equations of motion derive karne (proof) ke liye directly use hoti hai.
Worked example
Problem: A car accelerates uniformly from rest. Its `v-t` graph is a straight line passing through `(0, 0)` and `(10 s, 20 m/s)`. Find the distance it travelled in `10 s`.
- Line origin se `(10, 20)` tak hai, jo `x`-axis ke sath ek right-angled triangle banati hai.
- Displacement nikalne ke liye triangle ka area formula use karein (`1/2 * base * height`).
- Calculate karein.
Answer: `100 m`
Problem: A bus moves with a constant velocity of `15 m/s` for `4 s`. Find the distance travelled using its `v-t` graph.
- Constant velocity matlab horizontal line. Iske neeche rectangle banta hai.
- Displacement us rectangle ka area hoga.
Answer: `60 m`
Key points
- Area = Displacement: The area enclosed between the velocity-time graph line and the time axis gives the magnitude of the displacement.
- Triangle Area: For uniform acceleration starting from rest, the shape is a triangle. `Area = 1/2 * base * height`.
- Rectangle Area: For uniform velocity, the shape is a rectangle. `Area = length * width`.
Common mistakes
- Finding the slope instead of area: Exam me displacement pucha jata hai aur bacche jaldi mein `(y2 - y1) / (x2 - x1)` karke slope (acceleration) nikal dete hain. Hamesha dhyan rakhein ki kya pucha gaya hai.
- Forgetting the 1/2 in triangle area: Slanted line wale graph me bache `base * height` kar dete hain aur `1/2` multiply karna bhool jate hain, jisse answer double (galat) aata hai.
Recall questions
- What physical quantity is determined by the area under a velocity-time graph?
- If the v-t graph is a horizontal line, what shape will its area form?
- If the v-t graph is a slanted straight line starting from the origin, what is the formula to find the displacement?
Questions & answers
What does the area under the velocity-time graph represent?
It represents the magnitude of the displacement (or distance travelled) by the moving object.
Approach: Directly connect 'area under v-t graph' to displacement.
Calculate the distance travelled by a body moving with a constant velocity of 20 m/s for 5 s from its velocity-time graph.
100 m
Approach: The shape is a rectangle. Area = 20 * 5 = 100 m.
Continue learning
Previous: Interpret uniform acceleration on velocity-time graph
Previous: Calculate area of triangle and rectangle
Next: Apply s = ut + 1/2 at^2
Next: Apply 2as = v^2 - u^2