A train travelling at a speed of 100 m/s is decelerated to a speed of 50 m/s in a time of 15 s. Calculate the deceleration that the train is subjected to.
Answers
Answer:
ans
Explanation:
acc=v-u/t
acc=50-100/15
acc=-50/15
acc=-3.33m/s^2
Answer:
Motion Equations for Constant Acceleration in One Dimension
LEARNING OBJECTIVES
LEARNING OBJECTIVESBy the end of this section, you will be able to:
Calculate displacement of an object that is not accelerating, 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 , 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 Δt = tf−t0, taking t0 = 0 means that Δt = tf, 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, x0 is the initial position and v0 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, Δt=t. It also simplifies the expression for displacement, which is now Δx = x−x0. Also, it simplifies the expression for change in velocity, which is now Δv = v−v0. To , using the simplified notation, with the initial time taken to be zero,
\ \begin{cases}{\Delta}{t} &=& t \\{\Delta}{x} &=& x-{{x}_{0}}\\{\Delta}{v} &=& v-{{v}_{0}}\end{cases}
⎩
⎪
⎨
⎪
⎧
Δt
Δx
Δv
=
=
=
t
x−x
0
v−v
0
where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.