I need help doing these 4 problems. A car accelerates at a?

Question

Heres some problemms I need help with

1) A car a rest accelerates to a constant velocity of 10m/s. THe car then immediatly acclerates at 8.5m/s^2. What is its velocity agter it has gone 200m?


2) A car traveling at 7.5m/s accelerates at a constant of 9m/s^2 for a distance of 100m. How much time does it take?

3) A car in a driveway slips its brake and rolls into the street at a speed of 0.5m/s. If the car is acclerated at a constant of 1.0m/s How fast is it moveing 2 seconds after it reaches the street.

4)A Javelin leaves a throwers hand at a velocity of 21m/s at an angle of 43 degrees above the horizontal.
What is its maximum height?
How long does it take to reach that height?
How far along the ground does it travel?
What are its x and y displacements?

Answer 1

For 1, use the following equations:

distance=vi*t+.5*a*t^2
v=vi+a*t

where
vi=initial velocity 10 m/s
a=acceleration 8.5 m/s^2

200=10*t+.5*8.5*t^2
t=5.78 s (only the positive root is relevant)

v=10+8.5*5.78
=59.16 m/s

2. Same with different numbers
100=7.5*t+.5*9*t^2
t=3.95 s

3. More of the same
v=.5+1*2
v=2.5 m/s

4.This is more interesting.
There is a force acting on the javelin, gravity, which will cause the javeling to return to the ground.

It will follow the same equations, only now a bit of trig to look at the horizontal and vertical components separately.

First, the javelin will have a constant horizontal component of velocity is we ignore air friction:
vx=vi*cos(th)
vx=21*cos(43)
distance in the horizontal is
x=vx*t

For the vertical motion, as long as the javelin is above the ground, it will obey the following equation (again, ignoring air friction):

vy=vi*sin(th)-g*t
where g is the gravitational constant
9.81 m/s^2

y=vi*sin(th)*t-.5*g*t^2

the maximum height occurs when the vertical velocity is zero
or
t=vi*sin(th)/g
in this case,
t=1.46 s
the height is
y=vi*sin(th)*1.46-.5*g*1.46^2

=10.5 m

The distance traveled along the gound is
x=vx*t

at apogee,
x=22.4 m

it will hit the ground at twice that distance
44.8 m
j

Answer 2

1) This can only be answered if the 200 m is measured from the point at which the given acceleration starts. In that case, ∆V = √(2ax) = √(2*8.5*200) = 58.3095 m/s. Add the initial 10 m/s to get V = 68.3095 m/s.

2) d = Vot + .5at² → 100 = 7.5t + .5*9*t².....Solve this for t. I got 3.95sec

3) The ? is unreadable. Do you mean 1.0 m/s²? Do you mean 'BEFORE it reaches the street'? Is the car accelerating AFTER it reaches the street? NONE of these things are clear the way you have stated the Q.

4) a. 10.465 m
b. 1.4614 sec
c. 44.89 m
d. See a. & c.

Answer 3

I'm not too sure about this one but one approach would be to use the 4 equations of motion. There're relatively straight forward, I would write them but I can't figure out to write equations in this blasted box. They involve displacement, velocity (initial and final), acceleration and time.