At t0, one toy car is set rolling on a straight track with initial position 16.5

Question

At t0, one toy car is set rolling on a straight track with initial position 16.5 cm, initial velocity -2.8 cm/s, and constant acceleration 2.70 cm/s. At the same moment, another toy car is set rolling on an adjacent track with initial position 9.5 cm, initial velocity 4.80 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.) 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.) (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.) cm cm e) Explain the difference between question (a) and question (c) as clearly as possible

Explanation / Answer

a)

lets the velocity of toy car 1 be V1 , acceleration a1 = 2.7 cm/s , initial velocity U1 = - 2.8 cm/s

and similarly for toy car 2 are V2 , a2 = 0 , U2 = 4.8 cm/s

so lets say at time ta both the cars have same velocity , i.e. V1 = V2

from equations of motion we have V = U + at

U1 + a1 * ta = U2 + a2 * ta

U1 - U2 = ta * ( a2 - a1 )

ta = ( U1 - U2 )/( a2 - a1 )

ta = ( -2.8 - 4.8 ) / ( 0 - 2.7)

ta = 2.814 s

b)

substitute this time in the any of the velocity equation V1 or V2 because both will be same

V1 = -2.8 + 2.7 * 2.814 = 4.8 cm/s

V2 = 4.8 + 0 * 2.814 = 4.8 cm/s

4.8 cm/s is the ans

if you think logically , car 2 has no acceleration so its speed is constant , then that will be the speed of car 1 also when they both have same speeds.

c)

See from the given data , its understandable that car 1 has a -ve initial speed , which means it will travel backwards first , while car 2 is moving forward . typically it will be as shown below

car2(9.5)------------x2--------><----------x1----------(16.5)car1

lets say car1 moves x1 distance and car2 moves x2 distance where they pass each other

so it can be seen that x1 + x2 = 16.5 - 9.5 = 7 cm

we have equation of motion as

s = Ut + 1/2 at^2, let tc will be the time when they pass each other , this will be same for x1 and x2

x1 = U1 * tc + 1/2 a1 * tc^2

x2 = U2 * tc + 1/2 a2 * tc^2

x1 + x2 = ( U1 + U2 ) * tc + 1/2 ( a1 + a2 ) * tc^2 = ( -2.8 + 4.8 ) * tc + 1/2 ( 2.7 + 0 ) * tc^2 = 2 * tc + 1.35 * tc^2

7 = 1.35 tc^2 + 2 tc

solving this quadratic equation we gets roots as 1.628 and -3.18 but keeping these roots we cant get the answer because they are not converging

by trail and error we can find that at tc = 1.16 s

location of car 2 is 9.5 + 1.16 * 4.8= 15.068 cm

location of car 1 is 16.5 - ( -2.8 * 1.16 + 1/2 * 2.7 * 1.16^2) = 15.068 cm

t = 1.16 s ( bcoz of car 1 going back )

now find velocity at t = 1.16 for car 1

V = -2.8 + 2.7 * 1.16 = 0.332 cm/s

now both cars will cover same distance when they pass each other so

s1 = s2

V t + 1/2 a1 t^2 = V2 * t ( as a2 = 0 )

1.35 t^2 -4.468 t = 0

t = 3.309

so final will be tf = t + tc = 3.309 + 1.16 = 4.469 s

d)

1st location is 15.068 cm

the 2nd will be

initial position + distance travelled in t = 15.068 + 4.8 * 3.309

15.068 + 15.88 = 30.95 cm

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