Acar starts from rest and speeds up at a constant rate of 2.03 m/s^2 in a straight line until it reaches a speed of 21.1 m/s. the car then slows down at half the rate that it sped up until it stops. how far (in m) does the car move from start to stop?

The car moves from start to stop 328.88 m in total.

Explanation:

Vo= 0 m/s

V1= 21.1 m/s

V2= 0 m/s

a1= 2,03 m/s²

a2= -1.015 m/s²

Speed Up:

Speed up time:

V1= Vo + a1 * t1

t1= V1/a1

t1= 10.39 sec

total distance of speed up:

d1= Vo * t1 + (a1 * t1²)/2

d1= 109.57m

Slow Down:

V2= V1 - a2 * t2

t2= V1/a2

t2= 20.78 sec

total distance of slow down:

d2= V1 * t2 - (a2 * t2²)/2

d1= 219.31m

Total Distance:

TD= d1+d2= 109.57m + 219.31m

TD= 328.88 m

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

80cm/s.

24cm/s.

53.3cm/s.