Position, Velocity, Acceleration

[thaaampie]

A car is going 60mph (88ft/second) when it suddenly applies the brakes at time t=0 and skids (with constant deceleration) for 176 ft until it stops. What was the rate of deceleration? How long did it take before the car stopped?

An arrow was shot straight up from the ground at time t=0 seconds with an initial velocity of 49 m/sec. Assuming that the free-fall acceleration is -g=-9.8 m/sec^2, determine its maximum height, and when it hits the ground.

 

[Nex Vortex]

For the first problem you will want to use the following formula:
v_2^2=v_1^2+2ax
Remember that the car will eventually come to rest, so your final velocity will be 0 m/s. For the second part, you can use any other kinematic equation.

The second problem is very similar. At the maximum height, the velocity will be 0 m/s. The same formulas can be used. Just remember that when finding the time to hit the ground, the arrow must travel up and back down.

If you work the problems yourself and want somebody to check if they are right, or if you have more trouble with them, post back with your attempt at the solution and your answers and I will check them for you.

 

[HallsofIvy]

I would do those slightly differently from Nex Vortex. If the constant acceleration is -a, then the speed at time t is v(t)= v(0)- at= 88- at and the distance moved is x(t)= 88t- (a/2)t^2. Saying that the car stops after 176 feet means that v(t)= 88- at= 0 and that 88t- (a/2)t^2= 176 so that you have two equations to solve for a and t.

Similarly, for the second problem, the velocity of the arrow (positive upward) at time t is v(t)= 49- 9.8t and its height is x(t)= 49t- 4.9t^2. Set the second fomula equal to 0 to find the time it hit the ground. set first equation equal to 0 to solve for the time of its highest point (at that point it is no longer going up, so v(t) is not positive but it is not yet going down so v(t) is not negative) and put that t into the equation for x(t) to find the actual height.