A car travelling away from its starting point at a constant speed, then braking and coming to a stop. dar graph fast
Answers
Answer:
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
\[\text{Δ}t={t}_{\text{f}}-{t}_{0}\]
, taking
\[{t}_{0}=0\]
means that
\[\text{Δ}t={t}_{\text{f}}\]
, 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,
\[\text{Δ}t=t\]
. It also simplifies the expression for x displacement, which is now
Answer:
Determine the acceleration (in m/s2) of an object which ... .
moves in a straight line with a constant speed of 20.0 m/s for 12.0 seconds
changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds
changes its velocity from 0.0 mi/hr to 60.0 mi/hr in 4.20 seconds
accelerates from 33.4 m/s to 18.9 m/s over