Question #861

At t = 0, one toy car is set rolling on a straight track with initial position 16.5 cm, initial velocity -3.4 cm/s, and constant acceleration 2.30 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 9.5 cm, initial velocity 6.20 cm/s, and constant zero acceleration.

(a) At what time, if any, do the two cars have equal speeds? (Enter NA if the cars never have equal speeds.)

(b) What are their speeds at that time? (Enter NA if the cars never have equal speeds.)

(c) At what time(s), if any, do the cars pass each other? (If there is only one time, enter NA in the second blank. If there are two times, enter the smaller time first. If they never pass, enter NA in both blanks.)

(d) What are their locations at that time? (If there is only one position, enter NA in the second blank. If there are two positions, enter the smaller position first. If they never pass, enter NA in both blanks.)

(e) Explain the difference between question (a) and (c)

(a) At what time, if any, do the two cars have equal speeds? (Enter NA if the cars never have equal speeds.)

(b) What are their speeds at that time? (Enter NA if the cars never have equal speeds.)

(c) At what time(s), if any, do the cars pass each other? (If there is only one time, enter NA in the second blank. If there are two times, enter the smaller time first. If they never pass, enter NA in both blanks.)

(d) What are their locations at that time? (If there is only one position, enter NA in the second blank. If there are two positions, enter the smaller position first. If they never pass, enter NA in both blanks.)

(e) Explain the difference between question (a) and (c)

Expert's answer

The equation for the accelerated motion for the first car is v = v_{0} + at = -3.4 +2.3t

For the second:

v_{2} = v_{0} + at = 6.2 +0*t = 6.2.a) When their velocities are equal, we can write -3.4 +2.3t = 6.2 and find t.

2.3t = 9.6

t = 4.17 sec.

b) As the second car moves with constant speed, v_{1}=v_{2}=6.2 m/sec;

c) For the distance we have the following system of equations:

S_{1} = S_{01} + v_{01}t +at^{2}/2 = 0 - 3.4t + 2.3 t^{2}/2 = - 3.4t + 2.3 t^{2}/2;

S_{2} = S_{02} + v_{02}t +at^{2}/2 = 9.5 + 6.2 t;

S_{1} = S_{2}

_{ }- 3.4t + 2.3 t^{2}/2 = 9.5 + 6.2 t;

2.3 t^{2} -19.2 t - 19 = 0.

D = 543.44;

t_{1}= (19.2/2 +23.3)/2.3 = 14.3

t_{2} =(19.2/2 -23.3)/2.3 = -5.9

We accept only positive root,

thus the time when the two cars have equal speeds is 14.3 sec.

For the second:

v

2.3t = 9.6

t = 4.17 sec.

b) As the second car moves with constant speed, v

c) For the distance we have the following system of equations:

S

S

S

2.3 t

D = 543.44;

t

t

We accept only positive root,

thus the time when the two cars have equal speeds is 14.3 sec.

Learn more about our help with Assignments: MechanicsRelativity

## Comments

## Leave a comment