Question

The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Approximate the change in total stopping distance as speed changes from x = 25 to x = 26 miles per hour.

Answer 1

Answer:

%  change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

$T = 2.5 x + 0.5 x^{2}$

We will approximate this distance  using the relation

$f (x + dx) = f (x)+ f' (x)dx$

dx = 26 - 25 = 1

T' =  2.5 + x

Therefore

$f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x$

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× $25^{2}$ + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × $25^{2}$

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

%  change in stopping distance = $\frac{27.5}{375}$ × 100

%  change in stopping distance = 7.34 %