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sandra22
At t = 0, one toy car is set rolling on a straight track with initial position 13.5 cm, initial velocity -4.2 cm/s, and constant acceleration 2.60 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 8.5 cm, initial velocity 5.20 cm/s, and constant zero acceleration. (a) At what time, if any, do the two cars have equal speeds? (b) What are their speeds at that time? (c) At what time(s), if any, do the cars pass each other?
v=u+at, for first case equation becomes v=-4.2+2.6*t
for the second case equation becomes v1= 5.2, here a=0 there for when both toy car will have the same velocity then at that time v=v1 -4.2+2.6*t=5.2 therefore (2.6)t= (5.2+4.2) t= 9.4\2.6
t=3.7 approx, at t=3.7 seconds, both the toy car will have the same speed
hello ravi here initial velocity can be negative? I don't think so.
at the t = 3.7 seconds the speed of first toy car will be v=u+at= -4.2+(2.6)*(3.7) cm\sec
if it is negative then it is decelerating and 1st car will move back other than the second.
speed of second toy car will be the same as that of firsr
initial velocity is negative in the first case but acceleration is positive
so the velocity will come to zero, from negative and then will become positive, it means that the speed of direction of the body will change
formula of acceleration is a=v/t put values, 0.026=-0.042/t find the value of t from above equation, t=-1.615s And time can never never never be negative.
so the velocity will come to zero, from negative and then will become positive, it means that the speed of direction of the body will change if said above statement then how it becomes positive because it is not braking at any moment. Did you got my point?
u said that if velocity is negative then its decreasing but this concept is not true, negative means here that along the negative axis of coordinate system
oh dear. your logic is wrong this time. ok if you say that is on -ve x-axis. Then in contrast to it i say it is on -ve y-axis. I hope you got my point.
but i mean simply is that the negative sign of velocity does not says that velocity is decreasing
look. make it simple. At t = 0, one toy car is set rolling on a straight track with initial position 13.5 cm, initial velocity -4.2 cm/s, and constant acceleration 2.60 cm/s2 These lines are the condition for the first car. ok dear. If here velocity of first car is negative then how its acceleration could be positive. because both are relate by a=v/t. Let us consider for a moment that a is +ve and v is -ve then it implies that its time was -ve that in the formula both minuses are cancelled and we got +ve acceleration. can estimate this logic dear?
a=v\t is true only if u=o, but here u is not zero here a=v-u\t
oh yes now that's a point. I got you ravi623. That was helpful.
here is one more thing, if the velocity is negative and acceleration is positive then it means that first car was traveling with more negative value before the time t=0, so, we can consider that value here as initial velocity and initial velocity of firt car as final velocity then here (v-u) will be the positive quantity ,because here both u and v are negative but u is more than v in magnitude. So a= v-u\t will always be positive for the first case
So, what are the times when they pass each other?
Could one of you please clarify the final answer to this problem, writing it out so sandra22 can read it in one place? That would be great, thanks.
let us suppose that the car pass each other at time time t1 also suppose that both cars are moving along the x-axis of their respective coordinate systems and also the x-axis of the both system are parallel to each other and and origins of both coordinate systems coincide if we overlap one coordinate system upon each other So at the time t=0, first car is at the distance of 13.5 cm from the origin. and second car is at the distance of 8.5 cm from the origin
Still not a complete answer. Here we go then. If a body is moving long the x-axis, starting at initial position \( x_0 \), with initial velocity \( v_0 \) and accelerating at a constant rate of \( a \), then its position at time t, x(t), is given by \[ x(t) = x_0 + v_0t + \frac{1}{2}at^2 \] and it's velocity at time t, v(t) is given by \[ v(t) = v_0 + at \] Now we're told this about the first toy car: "one toy car is set rolling on a straight track with initial position 13.5 cm, initial velocity -4.2 cm/s, and constant acceleration 2.60 cm/s^2". Hence for that first car, \( x_0 = 13.5 cm \), \( v_0 = 4.2 cm/s \) and \( a = 2.60 cm/s^2 \). Therefore its velocity is \[ v_1(t) = 4.2 + 2.60t \] and position \[ x_1(t) = 13.5 + 4.2t + \frac{1}{2}2.60t^2 = 13.5 + 4.2t + 1.30t^2 \] Now, what about the second car: "another toy car is set rolling on an adjacent track with initial position 8.5 cm, initial velocity 5.20 cm/s, and constant zero acceleration." Hence its velocity, is \[ v_2(t) = 5.20 \] \[ x_2(t) = 8.5 + 5.2t \] That's the set up.
(a) At what time, if any, do the two cars have equal speeds? We have \[ v_1(t) = 4.2 + 2.6t \] and \[ v_2(t) = 5.2 \] The question is: when is \[ v_1(t) = v_2(t) \] i.e., when does \[ 4.2 + 2.6t = 5.2 \] \[ 2.6t = 1.0 \] \[ t = 1.0/2.6 = 0.38 sec \] Hence the two cars have equal velocity very quickly, t = 0.38 seconds after they start.
(b) What are their speeds at that time? As \( v_2(t) \) is constant for all time, the velocities of the two cars are equal when both of their velocities is 5.2 cm/sec.
(c) At what time(s), if any, do the cars pass each other? To find this set the two expressions for position equal to each other. That is, they are at the same position when \[ x_1(t) = x_2(t) \] \[ 13 + 4.2t + 1.3t^2 = 8.5 + 5.2t \] that is, \[ 1.3t^2 - t - 4.5 = 0 \] This is a quadratic equation in t. Its roots are \[ t = -1.51 \ sec \ \ \hbox{ and } t = 2.28 \ sec \] We're not interested in the negative time solution. Hence the two cars are the same position (and one is overtaking the other, which one is overtaking which?) when \[ t = 2.28 \ sec \]