A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passed the second point was 15.0 m/s. (a) What was the speed at the first point? (b) What was the magnitude of the acceleration? (c) At what prior distance from the first point was the car at rest? (d) Graph x versus t and v versus t for the car, from rest (t = 0).
What about part c and d????
Vi = 5m/s
a = 1.666
To answer these questions, we need to use the equations of motion for constant acceleration.
(a) To find the speed of the car at the first point, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the car started from rest, u = 0. We need to find the value of a. To calculate a, we can use the equation:
v = u + at
Rearranging the equation to solve for a:
a = (v - u) / t
Substituting the given values:
a = (15.0 m/s - 0) / 6.00 s = 2.50 m/s^2
Now we can calculate the initial velocity (speed) at the first point using the equation:
v = u + at
Rearranging the equation to solve for u:
u = v - at
Substituting in the known values:
u = 15.0 m/s - (2.50 m/s^2)(6.00 s) = 0 m/s
Therefore, the speed of the car at the first point is 0 m/s (it is at rest).
(b) The magnitude of acceleration is already calculated in part a and is equal to 2.50 m/s^2.
(c) To find the distance from the first point where the car was at rest, we need to use the equation for motion with constant acceleration where u = 0:
v^2 = u^2 + 2as
Rearranging the equation to solve for s:
s = (v^2 - u^2) / (2a)
Substituting in the known values:
s = (15.0 m/s)^2 / (2 * 2.50 m/s^2) = 22.5 m
Therefore, the car was at rest at a distance of 22.5 m from the first point.
(d) To graph x versus t (position versus time) and v versus t (velocity versus time) for the car, we can use the equations of motion.
For x versus t:
x = ut + (1/2)at^2
Since u = 0, the equation simplifies to:
x = (1/2)at^2
We can assume t = 0 as the starting point, so the equation becomes:
x = (1/2)at^2
For v versus t:
v = u + at
Since u = 0, the equation simplifies to:
v = at
Graphing these equations will give us a parabolic curve for x versus t and a linear line for v versus t.
I hope this helps you understand how to find the answers to these questions and how to graph the motion of the car! Let me know if you have any further questions.
d= vi*t+1/2 a t^2
60=6Vi+18a
but vf=vi+at
15=vi+6a
so you have two equations, two unknowns, solve. Multiply the second equation by 3, and subtract it frm the first is easy way.