Acar starts from rest with an acceleration of 2ms-2.at the same time a motor cycle starts with an uniform velocity of 20ms-1 in the same path 84m behind the car. how many times will the vehicles meet during their motion?

Twice

Explanation:

Use constant acceleration equation:

x = x₀ + v₀ t + ½ at²

where x is the final position,

x₀ is the initial position,

v₀ is the initial velocity,

a is the acceleration,

and t is time.

The position of the car is:

x = (84) + (0) t + ½ (2) t²

x = 84 + t²

The position of the motorcycle is:

x = (0) + (20) t + ½ (0) t²

x = 20t

When the positions are equal:

84 + t² = 20t

t² − 20t + 84 = 0

(t − 6) (t − 14) = 0

t = 6 or 14

The vehicles meet twice, once after 6 seconds, and once after 14 seconds.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer