A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.2 m/s. The car is a distance d away. The bear is 30 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely.

What is the maximum possible value for d?

Question

Erica Roach

Mathematics

18 May, 18:45

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.2 m/s. The car is a distance d away. The bear is 30 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely.

What is the maximum possible value for d?

+4

Answers (1)

Know the Answer?

Ellen · Answer 1

V = d/t

t = d/v

v (Bear) = 6.0 m/s, v (Tourist) = 4.2 m/s

(30 m + d) / 6 m/s = d / 4.2 m/s

4.2 (30 + d) = 6 d

126 + 4.2 d = 6 d

6 d - 4.2 d = 126

1.8 d = 126

d = 126 : 1.8 = 70 m

Answer:

The maximum possible value for d is 70 m.

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.2 m/s. The car is a distance d away. The bear is 30 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely.What is the maximum possible value for d?

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.2 m/s. The car is a distance d away. The bear is 30 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely.

What is the maximum possible value for d?