2 toy cars move horizontally toward each other. When they are 40m apart one has an initial velocity of 6m/s and acceleration of 4m/s^2 and the other one moves with constant velocity of 4m/s. 1. At what moment of time will they meet?
2. At what distance will they meet?
3. Draw in the same graph the x(t) for both cars.

Question

contestada

Respuesta :

aristocles aristocles · Answer 1 · 2019-10-13T08:18:36+02:00

Answer:

Part a)

[tex]t = 2.62 s[/tex]

Part b)

the distance moved by car 1 is 29.5 m and distance traveled by car 2 is 10.5 m

Explanation:

Part a)

As we know that car 1 is moving with speed v = 6 m/s and acceleration 4 m/s/s

Then car 2 is moving at constant speed 4 m/s

now the relative speed of two cars is

[tex]v_r = 6 + 4 = 10 m/s[/tex]

now the relative acceleration of two cars towards each other is given as

[tex]a_r = 4 m/s^2[/tex]

now we will have

[tex]d = v_i t + \frac{1}{2}at^2[/tex]

[tex]40 = 10 t + \frac{1}{2}(4)t^2[/tex]

[tex]t^2 + 5t - 20 = 0[/tex]

[tex]t = 2.62 s[/tex]

Part b)

In the above time distance traveled by the car which is moving at constant speed is given as

[tex]v = \frac{d}{t}[/tex]

[tex]4 = \frac{d}{2.62}[/tex]

[tex]d = 10.5 m[/tex]

so the distance moved by car 1 is 29.5 m and distance traveled by car 2 is 10.5 m

Part c)