Question

In the figure here, a red car and a green car move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at x_r = 0 and the green car is at x_g = 221 m. If the red car has a constant velocity of 23.0 km/h, the cars pass each other at x = 43.1 m. On the other hand, if the red car has a constant velocity of 46.0 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.

Answer 1

We have,

At t = 0s, x_r=0 and x_g = 221m.

If the red car has a constant velocity(v₁) of 23km/h = 6.38m/s and covers 43.1m distance(d₁).

Now, v₁=d₁/t₁

=> t₁=d₁/v₁

let t₁ be the time taken, then t₁ = 43.1/6.38 =6.755s

Now, If the red car has a constant velocity(v₂) of 46km/h = 12.77m/s and covers 76.6m distance(d₂).

Now, v₂=d₂/t₂

=> t₂=d₂/v₂

let t₁ be the time taken, then t₂ = 76.6/12.77 = 5.998s

We have the equation of motion, x-d = v₀t+at²/2

In the case when t = t₁ and d = 221m, x = 43.1m

43.1 - 221 = v₀(6.755) +a(6.755)²/2 .......(i)

In the case when t = t₂ and d =221m, x = 76.6m

76.6 - 221 = v₀(5.998) + a(5.998)²/2 .........(ii)

On solving equation (i) and (ii) for a and v₀, we get

v_o = -6.16m/s

and a = - 9.62m/s².

Hence, the intial velocity and acceleration of the green car is -6.16m/s and -9.62m/s² respectively.