JEE Main 2021 · 25 July · evening · Single correct · +4 / −1
The relation between time t and distance x for a moving body is given as t = mx² + nx, where m and n are constants. The retardation of the motion is : (When v stands for velocity)
- (a) 2 mv³
- (b) 2 mnv³
- (c) 2nv³
- (d) 2n²v³
Answer and solution
Answer (a) 2 mv³
Solution
Physics sir · 3 steps
Time is given as a function of x, so find dt/dx first: that is just 1/v.
- Velocity from t(x). Differentiate t = mx² + nx with respect to x. This gives dt/dx = 2mx + n. Since dt/dx = 1/v, we get v = 1/(2mx + n).
- Acceleration from v(x). Velocity is known as a function of x, so write acceleration as a = vdv/dx. Here dv/dx = -2m/(2mx+n)² = -2mv².
a = vdv/dx = v(-2mv²) = -2mv³
- So. The acceleration is negative, so the body is slowing down. Its magnitude, the retardation, is 2mv³. That is option (a).
The idea to keep. When time is given as a function of position, use dt/dx = 1/v, then a = vdv/dx. Constant terms in dt/dx drop out on the next differentiation.
Answer from the NTA final answer key. Solution by PrepWiser.