# Answer on Mechanics | Relativity Question for dk

Question #22713

At t = 0, one toy car is set rolling on a straight track with initial position 14.5 cm, initial velocity -2.5 cm/s, and constant acceleration 2.70 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 11.0 cm, initial velocity 5.00 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.)

1 s

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

2 cm/s

(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.)

3 s

4 s

(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.)

5 cm

6 cm

(e) Explain the difference between ques

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

1 s

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

2 cm/s

(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.)

3 s

4 s

(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.)

5 cm

6 cm

(e) Explain the difference between ques

Expert's answer

The equation for the velocityof the first car is

v1(t) = - 2.5 cm/s + 2.70 cm/s^2 * t

The equation for the velocity of the second car is

v2(t) = 5.0 cm/s + 0 * t = 5.0 cm/s = constant.

Let's find a moment of time T, when v1 = v2.

-2.5 cm/s + 2.70 cm/s^2 * T = 5.0 cm/s;

T = (5.0 + 2.5) cm/s / (2.70 cm/s^2) = 2.778 s.

Note that for this question, the initial positions of the cars do not matter.

Answer: 2.778 s.

v1(t) = - 2.5 cm/s + 2.70 cm/s^2 * t

The equation for the velocity of the second car is

v2(t) = 5.0 cm/s + 0 * t = 5.0 cm/s = constant.

Let's find a moment of time T, when v1 = v2.

-2.5 cm/s + 2.70 cm/s^2 * T = 5.0 cm/s;

T = (5.0 + 2.5) cm/s / (2.70 cm/s^2) = 2.778 s.

Note that for this question, the initial positions of the cars do not matter.

Answer: 2.778 s.

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