A car moves along an x axis through a distance of 900 m, starting at rest (at x

Question

A car moves along an x axis through a distance of 900 m, starting at rest (at x 0) and ending at rest (at x 900 m). Through the first of that distance, its acceleration is 2.25 m/s2. Through the rest of that distance, its acceleration is 0.750 m/s2. Its travel time through the 900 m is 56.6 s. Its maximum speed is 31.8 m/s.   

a. How long does the car take to reach the half of the distance?

b. estimate the speed of the car when it reaches half of the distance, 450m

c. find when the car moves at the speed of 23.0 m/s

d. find where the car is when it is moving at the speed of 23.0 m/s

Explanation / Answer

Total distance = D = 900 m

Distance through which car accelerates = d1 = (1/4) 900 = 225 m

a1 = acceleration = 2.25 m/s2

Distance through which car deaccelerate = d2 = 900 - 225 = 675 m

a2 = - 0.750 m/s2

a)

time taken to travel 225 m ::

d1 = Vi t + (0.5) a1 t2

225 = 0 (t) + (0.5) (2.25 ) t2

t = 14.14 sec

second half ::

speed reached at 225 m mark = V = 31.8 m/s

time taken to travel rest of 225 m distance ::

d2 = Vi t + (0.5) a2 t2

225 = (31.8) (t) + (0.5) (- 0.750 ) t2

t = 7.79 sec

Total time = 14.14 sec + 7.79 sec = 21.93 sec

b)

Using the formula ::

Vf = vi + at

Vf = 31.8 + (-0.750) (7.79) = 25.96 m/s

c)

Vf = Vi + at

23 = 0 + 2.25 t

t = 10.22 sec

d)

Vf2 = Vi2 + 2 a d

232 = 02 + 2 (2.25)d

d = 117.56 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.