Acar moves along an x axis through a distance of 980 m, starting at rest (at x = 0) and ending at rest (at x = 980 m). through the first 1/4 of that distance, its acceleration is + 4.20 m/s^2. through the next 3/4 of that distance, its acceleration is - 1.40 m/s2. what are (a) its travel time through the 980 m and (b) its maximum speed?

travel time is 32.4 s

maximum speed is 45.36 m/s

Explanation:

given data

distance = 980 m

acceleration = 4.20 m/s² for first 1/4 of that distance

acceleration = -1.40 m/s² for next 3/4 of that distance

to find out

travel time through the 980 m and maximum speed

solution

we know for first 1/4 of that distance is = = 245 m

so  equation of motion

s = ut + 0.5 ×at²     .............1

here u is initial speed = 0 and a is acceleration an t is time

s = ut + 0.5 ×at²

245 = 0+ 0.5 ×4.20 (t)²

t = 10.80 s

so

maximum speed at 1/4 of that distance

use equation of motion

v² - u² = 2as

put here value

v² - 0 = 2(4.20)× (245)

v = 45.36 m/s

so maximum speed is 45.36 m/s

and

for 3/4 distance

use equation of motion

v = u + at

here u is here 45.36 and a is acceleration and t is time and v final speed is 0

0 = 45.36 + (-1.40) × t

t = 32.4 s

so travel time is 32.4 s

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2