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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)

1 Answer

2 votes

Answer:

Step-by-step explanation:

a ) Let after time t , they have equal velocity .

v = u + at

for first toy car

v = - 3.4 + 2.3 t

velocity of second car after t = 6.2 cm /s because it has zero acceleration .

- 3.4 + 2.3 t = 6.2

t = 4.17 s .

b ) Their speeds are equal to 6.2 m /s each .

c ) Let after time t they pass each other , so their displacement will be equal

s = s₀+ ut + 1/2 at² for first car and s = ut for second car as for second a = 0

s = 16.5 -3.4t + 1/2 x 2.3 t² ., for second s = s₀+ 6.2 t

16.5 -3.4t + 1/2 x 2.3 t² = 9.5 + 6.2 t

1.15 t² - 9.6 t + 7 = 0

t = .8 s , 7.54 s .

d )

put t = .8 and 7.54 in the expression of s

s = 9.5 + 6.2 t

= 9.5 + 6.2 x .8

= 14.46 cm

and

9.5 +6.2 x 7.54

= 56.25 cm

e)

In case of a) speeds are equal of both the cars and in case of c ) their positions are equal .

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