JEE Main 2023 · 15 April · morning · Single correct · +4 / −1
The position of a particle related to time is given by x=(5 t²-4 t+5) m. The magnitude of velocity of the particle at t=2 s will be :
- (a) 14 ms⁻¹
- (b) 16 ms⁻¹
- (c) 10 ms⁻¹
- (d) 06 ms⁻¹
Answer and solution
Answer (b) 16 ms⁻¹
Solution
Physics sir · 3 steps
Velocity is the slope of position, so differentiate x once and then put in t = 2 s.
- Differentiate position. Instantaneous velocity is v = dx/dt. Differentiate each term of x = 5t² - 4t + 5. The constant 5 drops out.
v = d/dt(5t² - 4t + 5) = 10t - 4
- Substitute the time. At t = 2 s, v = 10 × 2 - 4 = 20 - 4 = 16 m s⁻¹. It is positive, so its magnitude is also 16 m s⁻¹.
- So. The particle moves at 16 m s⁻¹ at t = 2 s, which is option (b).
The idea to keep. When x is given as a function of t, differentiate once for velocity and twice for acceleration. Substitute the time only after you differentiate.
Answer from the NTA final answer key. Solution by PrepWiser.