Question

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.

Answer 1

Part a)

`t = 2.62 s`

Part b)

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

Step-by-step 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

`v_r = 6 + 4 = 10 m/s`

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

`a_r = 4 m/s^2`

now we will have

`d = v_i t + (1)/(2)at^2`

`40 = 10 t + (1)/(2)(4)t^2`

`t^2 + 5t - 20 = 0`

`t = 2.62 s`

Part b)

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

`v = (d)/(t)`

`4 = (d)/(2.62)`

`d = 10.5 m`

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

Part c)