Answer to Question #131994 in Mechanics | Relativity for Anika

Question

A boy starts walking down the street. He accelerates uniformly at 1.00 m/s^2 until he has walked 10 km. He then walks at a constant speed for 5 minutes. What is the total distance traveled by the boy?

Answer 1

When the body is moving with constant acceleration, the path depends on time as

$s = \frac{at^2}{2}$ (1)

and velocity at a certain point of time is

$v = at$ (2)

The time spent for passing 10km distance, according to (1), is

$t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2 \cdot 10^4}{1}} \approx 1.41 \cdot 10^2 = 141$ (s).

When the boy finished moving with the constant velocity, he started moving with the constant speed, the same as it was for the last period of time in 10km path, which can by found from (2):

$v = 1 \cdot 141 = 141$ (m/s).

Distance for the last part of path with constant velocity:

$s = vt = 141 \cdot 5 \cdot60 =42 \; 300 = 42.3$ km.

Total distance: $10+42.3 = 52.3$ km.

Please note: The conditions given in the task are far from reality, they are non-physical. The speed 141 m/s is 508 km/h, that is faster than any car created on Earth, and is rather the speed of plane. Obviously, the boy cannot have an acceleration 1m/s² along 10km path.