Question

A particle moving in a straight line with constant acceleration passes in succession A,B,C. The time taken from A to B is t and from B to C is 2t, AB = a, BC = b. Prove that the acceleration is (b-2a)÷(3t^2) and find the velocity at B. Assuming the initial velocity equals to u.

Answer 1

Answer of question #78002-Physics-Mechanics - Relativity

A particle moving in a straight line with constant acceleration passes in succession A,B,C. The time taken from A to B is t and from B to C is 2t, AB = a, BC = b. Prove that the acceleration is (b-2a)÷{3t^2} and find the velocity at B. Assuming the initial velocity equals to u.

Input Data:

Distance1: $AB = a$

Distance2: BC = b;

Time1: $t_1 = t$

Time2: $t_2 = 2t$

Acceleration: e;

Initial velocity: u;

Solution:

According to the equation of motion, the distances will be:

$$a = u t _ {1} + \frac {e t _ {1} ^ {2}}{2};$$ $$b = v t _ {2} + \frac {e t _ {2} ^ {2}}{2};$$

where $\pmb{v}$ is the speed reached at point $\pmb{B}$ :

$$v = u + e t _ {1} = u + e t;$$

Hence the distance $\pmb{b}$ after substitution of the initial data is equal to:

$$b = 2 t (u + e t) + \frac {4 e t ^ {2}}{2} = 2 u t + 4 e t ^ {2};$$

Distance $\pmb{a}$ is equal:

$$a = u t + \frac {e t ^ {2}}{2};$$

We multiply the distance $\mathbf{a}$ by 2 and subtract from distance $\mathbf{b}$ , as indicated in the condition. We obtain the following equation:

$$b - 2 a = 2 u t + 4 e t ^ {2} - 2 u t - e t ^ {2};$$ $$b - 2 a = 3 e t ^ {2};$$

Hence the acceleration is equal to:

$$e = \frac {b - 2 a}{3 t ^ {2}}$$

Answer:

Velocity at $B$:

$$v = u + et;$$

Acceleration:

$$e = \frac{b - 2a}{3t^2}, \quad \text{Q.E.D.}$$

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