At tequals0, a train approaching a station begins decelerating from a speed of 160 mi/hr according to the acceleration function a(t)equalsnegative 1280 (1 plus 4 t )Superscript negative 3 mi/hrsquared, where tgreater than or equals0. How far does the train travel between tequals0 and tequals0.2? Between tequals0.2 and tequals0.4?

The distance the train travels between 0 and 0.2 hours is 17.78 miles

The distance the train travels between 0.2 and 0.4 hours is 6.8376 miles

Step-by-step explanation:

v₁ = 160 mi/hr at t = 0

Deceleration a = 1280 ( 1+ 4t)^(-3)

The velocity at a future time is given by

Therefore, plugging the values we get

Which gives

Similarly, the position of the vehicle is given by

Which gives s(0) = 0 and

Therefore,

Which gives

Therefore, the distance traveled between points 0 and 0.2 is given by

Similarly the distance the train travels between 0.2 and 0.4 seconds is

.

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.