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A car 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?

Respuesta :

Answer:

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 = [tex]\frac{980}{4}[/tex] = 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

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