Question

A Toyota Camry and a student are at the same position. The driver of the Toyota Camry starts the vehicle from rest and moves with a constant acceleration of 3 m/s² along the x-direction. At the same time, the student starts walking in the y-direction at a constant speed of 1 m/s. What is the distance between the Toyota Camry and the student after 5 seconds? (10 marks)​

Answer 1

Step-by-step explanation:

For the car d = 1/2 a t^2 = 1/2 ( 3)(5^2) = 37.5 m

For the student d = 5s * 1 m/s = 5 m

these two distances are the legs of a right triangle

the distance between them will be found using the Pythagorean Theorem

d^2 = 37.5^2 + 5^2

d = 37.8 m