Question

Show all your work for each part of the question. The parts within the question may not have equal weight. Car A a0 Car B x=0 +x Figure 1 Two toy cars, A and B, are placed on parallel tracks as shown in Figure 1. At time $t = 0$ both cars are released from rest from position $x = 0$. Car A moves with constant acceleration of magnitude $a_0$ in the $+x$-direction and Car B moves with constant acceleration of magnitude $\frac{a_0}{2}$ in the $-x$-direction. (d) At time $t = t_0$ the cars A and B have traveled distances $d_A$ and $d_B$, respectively. Indicate whether $d_A$ is greater than, less than or equal to $d_B$. $d_A > d_B$_____ $d_A < d_B$_____ $d_A = d_B$_____ Justify your claim by using evidence from the graph sketched in part (c).

          Show all your work for each part of the question. The parts within the question may not have equal weight.
Car A
a0
Car B
x=0
+x
Figure 1
Two toy cars, A and B, are placed on parallel tracks as shown in Figure 1. At time $t = 0$ both cars are released from rest from position $x = 0$. Car A moves with constant acceleration of magnitude $a_0$ in
the $+x$-direction and Car B moves with constant acceleration of magnitude $\frac{a_0}{2}$ in the $-x$-direction.
(d) At time $t = t_0$ the cars A and B have traveled distances $d_A$ and $d_B$, respectively. Indicate whether $d_A$ is greater than, less than or equal to $d_B$.
$d_A > d_B$_____
$d_A < d_B$_____
$d_A = d_B$_____
Justify your claim by using evidence from the graph sketched in part (c).

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Show all your work for each part of the question. The parts within the question may not have equal weight.
Car A
a0
Car B
x=0
+x
Figure 1
Two toy cars, A and B, are placed on parallel tracks as shown in Figure 1. At time t = 0 both cars are released from rest from position x = 0. Car A moves with constant acceleration of magnitude a0 in
the +x-direction and Car B moves with constant acceleration of magnitude (a0)/(2) in the -x-direction.
(d) At time t = t0 the cars A and B have traveled distances dA and dB, respectively. Indicate whether dA is greater than, less than or equal to dB.
dA > dB
dA < dB
dA = dB
Justify your claim by using evidence from the graph sketched in part (c).

Added by Taylor F.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Km Neeraj

09/16/2021

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Step 1: The acceleration of car A is a0 and the acceleration of car B is a0/2. Show more…

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Show all your work for each part of the question. The parts within the question may not have equal weight. Car A a0 Car B x=0 +x Figure 1 Two toy cars, A and B, are placed on parallel tracks as shown in Figure 1. At time t = 0 both cars are released from rest from position = 0. Car A moves with constant acceleration of magnitude a0 in the+z-direction and Car B moves with constant acceleration of magnitude in the --direction. 20 2 (d) At time t = to the cars A and B have traveled distances da and dB, respectively. Indicate whether da is greater than, less than or equal to dB. da > dB da < dB dA = dB Justify your claim by using evidence from the graph sketched in part (c).

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Transcript

00:01 In this problem of motion in a straight line we have given a figure i'm showing you.

00:06 So this is the figure and car a and car b move in the same direction in adjacent lanes.

00:13 The position x of car a is given in figure.

00:18 The figure vertical scaling is set by x as equals to 32 meters.

00:23 Then this point is 20 meters, then this is 4 meters, 8 meters, 16 meters, this is 12 meters, then after 16 meters and then 20 meters and so on at t is equal to 0 car b is at x is equal to 0 so this this is the position of car b at t is equal to 0 with the velocity of initial velocity is equal to 12 meters per second for car b and a negative acceleration say this is ab this is the negative acceleration of car b which is represented as a b we have to find for part a what must be ab, this is ab such that the cars are momentarily side by side at t is equal to 4 second so first we have to find the position of car a this is for car a now at 4 second this is at here 1 2 3 and 4 second position is at this is 28 meters so for car a a position at at t is equal to 4 second is x is equal to this is 28 meters so this is 28 meters now for car b it should be again at 28 meters so we say that this is for car b equation would be 28 meters is equal to this is initial velocity is 12 meters so this is 12 multiplied with 4 plus half multiplied with 4 plus half multiplied with this this is acceleration and t square.

02:21 This is 4 square.

02:22 From here we say that ab is equals to minus 2 .5 meters per second square which is the answer for part a.

02:32 Now this is the answer for part a...

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