A car is traveling at a constant speed of 31.8 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.58 km away?
It takes 2580/31.8 = 81.13 seconds for car 1 to get to the exit.
So, for car 2,
1/2 a*81.13^2 = 2580
just solve for a.
To find the acceleration of the second car, we can use the kinematic equations of motion. We have the following information:
Initial velocity of the second car (u) = 0 m/s (since it starts from rest)
Distance to be covered (s) = 2.58 km = 2.58 * 1000 m (converting km to m)
Final velocity of the second car (v) = 31.8 m/s (since it meets the first car at the next exit)
Acceleration of the second car (a) = ?
We can use the equation:
v^2 = u^2 + 2as
Rearranging the equation to solve for acceleration:
a = (v^2 - u^2) / (2s)
Plugging in the values, we get:
a = (31.8 m/s)^2 / (2 * 2.58 * 1000 m)
Calculating this expression:
a = 1011.24 m^2/s^2 / (5160 m)
Simplifying the units:
a = 0.196 m/s^2
Therefore, the second car must maintain an acceleration of 0.196 m/s^2 in order to meet the first car at the next exit.