Un automobil care se misca uniform accelerat parcurge distanta de 60 m dintre doua puncte in 6 secunde. viteza sa in momentul in care trece prin cel de-al doilea punct este de 15m/s. care este acceleratia automobilului si la ce distanta inainte de primul punct a fost in repaus?

consider the motion between first and second point

v₁ = velocity at first point

v₂ = velocity at second point = 15 m/s

a = acceleration of the car

t = time of travel between the two points = 6 sec

using the kinematics equation , velocity at second point is given as

v₂ = v₁ + a t

15 = v₁ + 6 a

v₁ = 15 - 6 a                                          eq-1

d = distance traveled between two points = 60 m

using the kinematics equation

d = v₁ t + (0.5) a t²

inserting the values and using eq-1

60 = (15 - 6 a) (6) + (0.5) a (6)²

a = 1.67 m/s²

using eq-1

v₁ = 15 - 6 a

v₁ = 15 - 6 (1.67)

v₁ = 5 m/s

D = distance traveled before point one

v₀ = initial velocity of car at the point of start = 0 m/s

Using the kinematics equation

v²₁ = v²₀ + 2 a D

5² = 0² + 2 (1.67) D

D = 7.5 m