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

Answer:

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

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

$d = v_i t + \frac{1}{2}at^2$

$40 = 10 t + \frac{1}{2}(4)t^2$

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

$t = 2.62 s$

Part b)

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

$v = \frac{d}{t}$

$4 = \frac{d}{2.62}$

$d = 10.5 m$

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

Part c)