A car goes from 100 m/s to a full stop for in 9.5 seconds. What is the acceleration?

Calculate displacement of an object that is not acceleration, given initial position and velocity.

Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.

Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.

Four men racing up a river in their kayaks.

Figure 1. Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr).

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.

Notation: t, x, v, a

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is

Δ

t

=

t

f

−

t

0

, taking

t

0

=

0

means that

Δ

t

=

t

f

, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is,

x

0

is the initial position and

v

0

is the initial velocity. We put no subscripts on the final values. That is,

t

is the final time,

x

is the final position, and

v

is the final velocity. This gives a simpler expression for elapsed time—now,

Δ

t

=

t

. It also simplifies the expression for displacement, which is now

Δ

x

=

x

−

x

0

. Also, it simplifies the expression for change in velocity, which is now

Δ

v

=

v

−

v

0

. To summarize, using the simplified notation, with the initial time taken to be zero,

Explanation:

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.