Car acceleration to meet at exit 1.8 km away and bus catch-up problem

[Kadaj]

I can't seem to figure these problems if anyone can give me the concept and start me off that would be great.

Thanks in advance

Problem one: A car is traveling at a constant speed of 27 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 1.8 km away?

Problem two: A bus stop to pick up riders. A woman is running at a constant velocity of +5.0 m/s in an attempt to catch the bus When she is 11 m from the bus, it pulls away with a constant acceleration of +1.0 m/s^2. From this point, how much time does it take her to reach the bus if she keeps running with the same velocity?

*edit*
Oh sorry i posted this after i read the before you post thread Once again I am sorry for posting Homework help here

 

[Pyrrhus]

For the first problem consider the formulas

$$vt = x$$

$$\frac{1}{2}at^2 = x$$

You go the speed for the car and the distance x they will both travel, what can you do with that?

For the second problem

She will do

$$vt = x + 11$$

while the bus will do

$$\frac{1}{2}at^2 = x$$

 

[wais]

No problem, I am here to assist you with these problems. For problem one, we need to use the equation d = vt + 1/2at^2, where d is the distance traveled, v is the initial velocity, a is the acceleration, and t is the time. We know that the first car is traveling at a constant speed of 27 m/s, so v = 27 m/s. We also know that the second car starts from rest, so its initial velocity is 0 m/s. We can rearrange the equation to solve for a: a = 2(d - vt)/t^2. Plugging in the values, we get a = 2(1.8 km - 27 m/s * t)/(t^2). Since we want the two cars to meet at the next exit, we know that the distance traveled by both cars must be the same. Therefore, we can set the equations for both cars equal to each other and solve for t. This will give us the time it takes for the two cars to meet. Once we have t, we can plug it back into the equation for a to find the acceleration needed for the second car.

For problem two, we can use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time. We know that the woman is running at a constant velocity of +5.0 m/s, so v0 = 5.0 m/s. The bus is accelerating at +1.0 m/s^2, so a = 1.0 m/s^2. We can rearrange the equation to solve for t: t = (v - v0)/a. Plugging in the values, we get t = (11 m - 5.0 m/s)/(1.0 m/s^2). This gives us the time it takes for the woman to catch up to the bus.