Physics help, please!

Question

This is my first year taking physics, and since my teacher hasn't covered this material yet, I'm having a hard time with these two problems. Can someone please explain to me how to solve them? Thanks in advance!

A car is traveling at a constant speed of 18 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 1.7 km away? (in meters/s/s)

A ball is thrown straight upward and rises to a maximum height of 40 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?

(in meters)

Answer 1

idk

Answer 2

<< A car is traveling at a constant speed of 18 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 1.7 km away? (in meters/s/s) >>

Let

T = time for the car (doing constant speed) to reach the next exit

Distance = Velocity * Time

OR

D = VT

From the given of the problem, D =1.7 km = 1700 meters and V = 18 m/sec.

Therefore, substituting values,

1700 = 18(T)

Solving for T,

T = 1700/18 = 94.44 seconds

NOTE -- it should take the second car the same time to reach the next exit in order for them to meet. The working equation is

D = VoT + (1/2)aT^2

where

D = 1700 m

Vo = initial velocity = 0 (car starts from rest)

T = time = 94.44 sec

a = acceleration

Substituting appropriate values,

1700 = 0 + (1/2)(a)(94.44^2)

Solving for "a",

a = 2 * 1700/(94.44^2)

a = 0.381 m/sec^2

<< A ball is thrown straight upward and rises to a maximum height of 40 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?

(in meters) >>

Your working equation is

Vf^2 - Vo^2 = 2gs

where

Vf = velocity of ball at its maximum height = 0

Vo = initial velocity

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

s = maximum height attained = 40 m

Substituting appropriate values,

0 - Vo^2 = 2(-9.8)(40)

NOTE the negative sign attached to the acceleration due to gravity. This merely implies that the ball is slowing down as it is going up.

Solving for Vo,

Vo = 28 m/sec.

Use the same formula again for the second part of the problem,

Vf^2 - Vo^2 = 2(g)(h)

where

Vf = (1/2)Vo = (1/2)28 = 14 m/sec.

h = height above launch point where Vf = 14 m/sec.

and all other terms are previously defined.

Substituting values,

14^2 - 28^2 = 2(-9.8)h

Solving for "h",

h = (14^2 - 28^2)/(2 * -9.8)

h = 30 meters