1)At the instant traffic turns green, an automobile that has been waiting at an intersection of roads accelerats with 5ms-2. At the same instant a truck traveling with a constant velocity. (i) how far beyond its starting point will be automobile overtake the truck? (ii)how fast will it be traveling? (2) An object is projected with an initial velocity U at an angle B with a plane inclinee at an angle A with the horizontal. Derive an expression for the angle of the projectile.

x1 = 1/2 at^2

x2 = vt

Obviously we need the speed of the truck.

To solve these questions, we will break them down step by step and explain the necessary formulas and concepts along the way.

1) (i) How far beyond its starting point will the automobile overtake the truck?

To solve this, we need to calculate the time it takes for the automobile to catch up with the truck. Once we have the time, we can find the distance using the formula: distance = initial velocity × time + 0.5 × acceleration × time squared.

To find the time, we need to consider that both the automobile and the truck are moving from the same instant and have different initial conditions. Initially, both vehicles start at the same point.

Let's assume U_a is the initial velocity of the automobile and v_t is the speed of the truck (constant velocity).

The equation for the distance traveled by the automobile is: distance_a = U_a × time + 0.5 × 5ms^(-2) × time^2.

The equation for the distance traveled by the truck is: distance_t = v_t × time.

To determine when the automobile overtakes the truck, we need to set the distances equal to each other and solve for time: U_a × time + 0.5 × 5ms^(-2) × time^2 = v_t × time.

Simplifying the equation, we get: 0.5 × 5ms^(-2) × time^2 + (U_a - v_t) × time = 0.

We can solve this quadratic equation for time using the quadratic formula.

(ii) After finding the time, to determine the speed of the automobile at the moment of overtaking, we can use the formula: final velocity = initial velocity + acceleration × time.

2) An object is projected with an initial velocity U at an angle B with a plane inclined at an angle A with the horizontal. We need to derive an expression for the angle of the projectile.

To derive the expression, we can break down the initial velocity into its horizontal and vertical components. Let's call the horizontal component U_x and the vertical component U_y.

U_x = U × cos(B) (horizontal component)
U_y = U × sin(B) (vertical component)

Now, considering the plane inclined at an angle A, the vertical component U_y will be affected by gravity. The gravitational force acts in the downward direction and can be considered as -g, where g is the acceleration due to gravity.

To find the angle of the projectile (let's call it C), we can determine the ratio of the horizontal and vertical components as:

tan(C) = (U_y - (-g)) / U_x.

Simplifying the expression, we get:

tan(C) = (U × sin(B) + g) / (U × cos(B)).

By using the trigonometric identity tan(C) = sin(C) / cos(C), we can rewrite the equation as:

tan(C) = (sin(B) + g/U) / cos(B).

Finally, we can solve for C by taking the inverse tangent of both sides of the equation:

C = tan^(-1)((sin(B) + g/U) / cos(B)).

This expression gives us the angle C, which represents the angle of the projectile with respect to the inclination of the plane.