(a) The particle is momentarily stopped when its velocity is zero. Its velocity as a function of time is the derivative of its coordinate with respect to time.
(b) Substitute the value for the time of stopping into the function given for the coordinate.
(c) and (d) Put 
equal to zero in the given function and solve for 
. The equation is quadratic in 
and so has two solutions, one positive and one negative.
(f) Suppose the function is 
, where 
is a constant. This is the same as the original function except it is shifted in time by 
. if 
is positive it is shifted in the positive 
direction and if 
is negative it is shifted in the negative 
direction. The second function is identical to 
as you can easily see by squaring the quantity in parentheses.
(g) Set the derivative of the new function equal to zero and solve for 
, then substitute back into the function to find the coordinate at which the particle stops.
[(a) 0; (b) 
; (c) 
; (d) 
; (f) 
; (g) increase]
. To calculate the particle's coordinate at the beginning of the interval substitute 
into the equation for 
and to calculate its coordinate at the end of the interval substitute 
. 
, of course, is the difference and 
is 
. Do the same for 
and for 
.
by taking the average of the two coordinates found in part (a), then solve 
for 
. Finally, substitute this value for 
into the expression for the velocity as a function of time (the derivative of the coordinate).
; (b) 
; (c) 
; (d) 
; (e) 
]
.
. Find the coordinate at the beginning and end of the interval and subtract the former from the latter to get 
.
; (b) 
; (c) 
; (d) 
; (e) 
; (f) 
; (g) 
; (h) 
; (i) 
]
, where 
is the electron's velocity at any time 
, 
is its velocity at time 
, and 
is its acceleration. Let 
at the instant the electron's velocity is 
and evaluate the expression for (a) 
and for 
.
; (b) 
]
is given by 
, where 
is its coordinate at 
, 
is its velocity then, and 
is its acceleration. According to the graph 
. The slope of the curve at 
appears to be zero so you may take 
to be zero. Thus 
. The graph also shows that 
at 
. Put this information into the equation and solve for 
.
; (b) 
]
axis at the position of train A when it starts slowing and suppose the train has velocity 
(
from the graph) at that time. Its velocity as a function of time is given by 
, where 
is its acceleration. This is the slope of the upper line on the graph and is negative. Solve for the time when train A stops and use 
to find its position when it stops.
, where 
is its velocity at 
(
from the graph), and 
is its acceleration. This is the slope of the lower line on the graph and is positive. Solve for the time when train B stops and use 
to find its position when it stops. Here 
. The separation of the trains when both have stopped is the difference of the coordinates you have found.
]
axis of a coordinate system so the origin is at the front end of the passenger train when it starts slowing and suppose the train is moving in the positive 
direction. The coordinate of the front end is then given by 
, where 
is its velocity at 
and 
is its acceleration. Its velocity is given by 
. The coordinate of the back end of the locomotive is given by 
, where 
is its velocity.
equal to the expression for 
and the expression for 
equal to the expression for 
. Solve these equations simultaneously for 
. (You can also solve for the time when these conditions are met but you are not asked for this.) Be sure to convert the given speeds to meters per second.
]
, where 
is the velocity with which it is thrown, 
is the height of the bridge above the roadway, and 
(
) is the time it is thrown. Here the origin is placed at the roadway and 
is taken to be positive in the upward direction. You need to know the value of 
.
. Set 
equal to zero and solve for 
using 
. Use this value and 
in the equation for 
and solve for 
.
]
, where 
is the time the drop started. Suppose 
for the first drop, 
for the second, 
for the third, and 
for the fourth. Each drop hits the floor when 
, the height of the nozzle above the floor. For the first drop 
. Solve for 
. The coordinate of the second drop at this time is 
and the coordinate of the third drop is 
.
; (b) 
]
when falls past that point. If 
is the top-to-bottom dimension of the window, then 
, where 
is the time to pass the window. Solve for 
.
. To find the length of the second segment solve for the distance the ball falls in 
, starting with a downward velocity of 
. The time here is the time for the ball to pass the window plus the time for it to fall from the bottom of the window to the sidewalk.
]
]
be the initial velocity and 
the time to reach the highest point. The velocity at the highest point is zero. Solve 
for 
, then use 
to compute the coordinate 
of the highest point, relative to the top of the building. Lastly, use the same equation, but with 
to compute the coordinate of the ground relative to the top of the building. The magnitude of the coordinate is the height of the building.
; (b) 
; (c) 
]
for the acceleration 
. Here 
is the final velocity (0) and 
is the initial velocity (
). You need to convert this to meters per second. Solve 
for the time of braking 
. The distance traveled during the reaction time 
is 
.
; (b) 
; (c) 
; (d) 
; (e) braking; (f) 
]
to compute the change in the coordinate. The average acceleration is the change in the velocity divided by the time for the change. Differentiate the expression for 
to obtain an expression for the velocity, then evaluate it for the two times. The instantaneous acceleration is the derivative of the velocity with respect to time.
; (b) 
; (c) 
; (d) 
; (e) 
; (f) 
]
, where 
is the velocity of the stone when it is at A, 
is its velocity when it is at B, and 
is the height of B above A. Put 
equal to 
and 
, then solve for 
. 
to compute the maximum height 
that the stone rises above point B.
; (b) 
]
; (b) 
]
; (b) 
]
to find the distance traveled at each rate, then sum the distances to find the total distance. for part (b) use 
to find the time for each rate, then sum the times to find the total time.
; (b) 
(c) 
; (d) 
]
, with 
to find an expression for the time to reach the ground and 
to find an expression for the velocity of the ball just before it hits the ground. Here 
is a speed, not a velocity. For parts (c) and (d) use 
and 
.
; (b) 
; (c) same as (a); (d) 
, greater]