Question

(10%) Problem 9: Two cars are heading towards one another. Car A is moving with an acceleration of a 4-5 ms. Car B is n with an acceleration of aB 9 m/s. The cars are initially at rest and separated by a distance d- 2300 m on the x-axis. oving 38 25% Part (a) At what time, in seconds, do the cars meet? Grade Summary 7.69 Potential 100% 7 8 9 HOME coso asino atan acotan0 sinho cosho tan cotanho sino tan ( Attempts remaining: (0% per attempt detailed view cotano coso 0% END . Degrees Radians BACKSPACE DELI CLEAR Submit Hint I give up! Hits: 0% deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback. Submission History All Date tines are displayed in Central Stodord Time Red sodulssion date tiwes inolcate lote work Date Time Answer Hints Feedback I Jan 21, 201 8:24 PM 25% Part (b) What is the displacement. În meters, of Car A? 曇▲ 25% Part (c) What is the displacement in meters, of Car B? 25% Part (d) Now assume that both cars start with an initial velocity. Car A's initial velocity is VOA 23 ms and Car Bs initial velocity is "B

Answer 1

Part a)

We have following information

u_A = 0m/s (at t =0, it is at rest)

u_B = 0m/s (at t =0, it is at rest)

a_A = 5m/s²

a_B = -9 m/s²

Both cars are travelling in opposite direction ( a_B is negative and a_A is positive) and we will denote the time when they will meet as "t"

Distance covered by each car will be given as:

s_A = u_A t + (1/2) a_A t² = 0 * t + (1/2) * 5 t² = 2.5t²

s_B = u_B t + (1/2) a_B t² = 0 * t - (1/2) * 9 t² = -4.5t²

As both cars are travelling in opposite direction and separated by d = 2300m; when they will meet will have

s_A - s_B = d

2.5t² - (-4.5t²) = 2300

7 t² = 2300

t = SQRT (2300 / 7) = 18.13 second

Part b)

Displacement of car A

s_A = 2.5t² = 2.5 * 18.13² = 821.4 m

Part c)

Displacement of car B

s_B = -4.5t² = -4.5 * 18.13² = -1478.6 m