A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.65 m/s. The car is a distance d away. The bear is 32.4 m behind the tourist and running at 5.68 m/s. The tourist reaches the car safely. What is the maximum possible
value for d?

146.27 m

Explanation:

From the question given above, the following data were obtained:

Velocity of tourist (vₜ) = 4.65 m/s

Distance travelled by tourist (dₜ) = d

Velocity of bear (v₆) = 5.68 m/s

Distance travelled by bear (d₆) = 32.4 + d

Value of d =?

Next, we shall determine the time taken for tourist and the bear to get to the car.

For the tourist:

Velocity of tourist (vₜ) = 4.65 m/s

Distance travelled by tourist (dₜ) = d

Time (tₜ) =?

vₜ = dₜ / tₜ

4.65 = d/tₜ

Cross multiply

4.65 × tₜ = d

Divide both side by 4.65

tₜ = d / 4.65

For the bear:

Velocity of bear (v₆) = 5.68 m/s

Distance travelled by bear (d₆) = 32.4 + d

Time (t₆) =?

v₆ = d₆ / t₆

5.68 = (32.4 + d) / t₆

Cross multiply

5.68 × t₆ = (32.4 + d)

Divide both side by 5.68

t₆ = (32.4 + d) / 5.68

NOTE: Both the tourist and the bear have the same time. Thus, to obtain the value of d, we shall equate both the time taken for the tourist and the bear together. This is illustrated below:

Time taken by tourist (tₜ) = time taken by the bear (t₆)

tₜ = d / 4.65

t₆ = (32.4 + d) / 5.68

tₜ = t₆

d / 4.65 = (32.4 + d) / 5.68

Cross multiply

d × 5.68 = 4.65 (32.4 + d)

5.68d = 150.66 + 4.65d

Collect like terms

5.68d – 4.65d = 150.66

1.03d = 150.66

Divide both side by 1.03

d = 150.66 / 1.03

d = 146.27 m

Thus, the maximum value of d is 146.27 m

high-sugar diet and high cholesterol