Apply v = u + at
Scenario
Aapki flight runway par rest se start hoti hai aur lagatar accelerator full rakhti hai. Aapko ye pata karna hai ki `15 seconds` baad take-off ke time uski speed kya hogi.
Time, acceleration, aur starting speed ko use kar ke exactly future ki speed kaise predict karte hain?
Mental model
Final Speed = Shuruati Speed + (Speed badhne ka rate × Kitni der tak badhi).
Ye equation bas common sense ka math translation hai. `at` wala part ye batata hai ki total speed me kitna 'izaafa' (increase/decrease) hua. Us izaafa ko purani speed `u` me jod do, aur aapko nayi speed `v` mil jayegi.
Explanation
Newton ki first equation of motion hai: `v = u + at`. Darasal, ye equation directly acceleration ke definition wale formula (`a = (v-u)/t`) se hi aati hai. Jab aap time `t` ko left mein bhej kar `a` ke sath multiply karte hain, aur phir `u` ko shift karte hain, to ye nayi shakal le leti hai.
Is equation ka sabse bada faida tab hota hai jab problem mein 'displacement (`s`)' ka koi zikr na ho. Agar aapko ek gaadi ki initial speed pata hai (`u`), acceleration pata hai (`a`), aur ye pata hai ki wo kitni der tak accelerate ki (`t`), toh aap easily bata sakte hain ki us specific time ke end mein gadi kis speed par hogi (`v`).
Signs ka dhyaan rakhna yahan bohot zaruri hai. Agar object breaks maar raha hai ya oppose ho raha hai, toh acceleration `a` negative hogi. Aise mein (`at`) wala term negative ho jayega, aur final speed `v` aapki initial speed `u` se kam aayegi. Ye equation object ke stop hone ka exact time nikalne ke liye bhi perfect hai, kyunki jab object stop hota hai toh uski final velocity `v = 0` hoti hai.
Ek tip: Humesha dhyaan rakhein ki `u` aur `v` dono same units (prefer `m/s`) mein hon, aur time bhi unhi ke matching units (prefer `s`) mein ho. Agar nahi hain, toh solve karne se pehle conversion rule zaroor lagayein.
Worked example
Problem: A car starts from rest and accelerates uniformly at a rate of `4 m/s^2`. What will be its velocity after `5 seconds`?
- Starts from rest means `u = 0`, `a = 4`, `t = 5`. Pucha hai final velocity (`v`).
- First equation use karein.
- Multiply and add.
Answer: `20 m/s`
Problem: A cyclist moving at `10 m/s` applies brakes and produces a retardation of `2 m/s^2`. Find his final velocity after `3 seconds`.
- Initial velocity `u = 10`. Retardation means `a = -2`. Time `t = 3`.
- Substitute these values into `v = u + at`.
- Calculate the result.
Answer: `4 m/s`
Key points
- First Equation of Motion: `v = u + at`. It establishes a relationship between velocity, acceleration, and time without needing distance.
- Deceleration cases: If brakes are applied, use a negative value for `a`, which will naturally make the final velocity `v` smaller.
- Finding stopping time: If an object stops, set `v = 0` and solve for `t = -u / a`.
Common mistakes
- Forgetting the negative sign for brakes: Bache retardation dekhte hain par equation me `v = u + at` me `a` ki jagah seedha `+2` daal dete hain, jisse answer badh kar `16` aa jata hai. Brakes hamesha speed ghatate hain.
- Not converting units first: Agar speed `36 km/h` me hai aur time `2 s` me, to bina `5/18` use kiye directly `36 + (a*2)` karne par numerical pura fail ho jayega.
Recall questions
- State the first equation of motion.
- Which variable is completely missing from the first equation of motion?
- What is a key takeaway about apply v u at?
Questions & answers
A train starting from rest attains a velocity of `72 km/h` in `5 minutes`. Assuming that the acceleration is uniform, find the acceleration.
`1/15 m/s^2`
Approach: Starts from rest -> `u = 0`. `v = 72 * 5/18 = 20 m/s`. `t = 5 * 60 = 300 s`. Use `v = u + at` -> `20 = 0 + a*300`. `a = 20/300 = 1/15 m/s^2`.
A particle moves with an initial velocity of `5 m/s` and a constant acceleration of `2 m/s^2`. Find its velocity after `10 s`.
`25 m/s`
Approach: Direct application of `v = u + at`. `v = 5 + (2 * 10) = 5 + 20 = 25 m/s`.
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