Question

A car 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?

Answer 1

Twice

Step-by-step 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.