Moving with an initial speed of 54 km / h, the car covered 155 m in 10 seconds.

Moving with an initial speed of 54 km / h, the car covered 155 m in 10 seconds. With what acceleration did the car move and what speed did it acquire at the end of the journey?

Equally accelerated rectilinear movement of the car on a certain section of the path, length S, is described by the equations:

S = v₀ ∙ t + a ∙ (t ^ 2) / 2 and v = v₀ + a ∙ t, where v₀ is the initial speed of movement, v is the final speed of movement in a given section, a is acceleration, t is the time of movement along a given section of the path.

Then:

a = 2 ∙ (S – v₀ ∙ t) / (t ^ 2).

From the condition of the problem it is known that the car, moving with an initial speed of v₀ = 54 km / h = 15 m / s in time t = 10 s, covered the path S = 155 m. We get:

a = 2 ∙ (155 m – 15 m / s ∙ 10 s) / (10 s) ^ 2;

a = 0.1 m / s ^ 2.

At the end of the considered path, the car acquired speed:

v = 15 m / s + 0.1 m / s ^ 2 ∙ 10 s;

v = 16 m / s.

Answer: the car was moving with an acceleration of 0.1 m / s ^ 2 and at the end of the path it acquired a speed of 16 m / s.