how to find distance with velocity and time

In this commodity, we will larn how to summate distance from acceleration and velocity. Nosotros would consider a special case of motility where our object under consideration is moving with constant acceleration. Here in this article, we volition simply discuss the formula for calculating the distance from acceleration and velocity only. The formula we would be using does not involve fourth dimension. Nevertheless, if you lot are aware of initial and final velocity and dispatch y'all can find time using the first equation of motility which is $v=u+at$.

Before going any farther we must be aware of all the terms used. And so,

Altitude:- Distance covered by a moving object refers to how much footing the object has covered without any regard to the direction of motion. SI unit of measurement for measuring distance is a meter.

Velocity:- Distance traveled by the moving body per unit of time gives the measure of the velocity of the object. It tells nigh how far an object moves in a given interval of time. SI unit for measuring velocity is meter per second (one thousand/s).

Acceleration:- Acceleration is the rate of change of velocity of an object with respect to time. Like velocity, it is a vector quantity having both magnitude and dispatch. Its unit is $m/s^2$

Calculate Altitude from Acceleration and Velocity

Since nosotros are because the motion of the body or object with constant acceleration we have an equation of motion that relates all these quantities. This equation is the third equation of motion and is given by the relation

$v^2=u^2+2as$

Here,

$s$ – is the distance traveled by the object in time $t$

$u$ – is the initial velocity

$v$ – is the final velocity and

$a$ – is the constant dispatch of the moving object

You can use this formula in various situations involving distance, initial velocity, final velocity and acceleration motion. You need to take a knowledge of 3 quantities to find the 4th quantity. It must be noted that this equation does not involve the time interval of motility of the object.

Solved Example

This solved example shows How to Calculate Time and Altitude from Acceleration and Velocity

Example 1

A torso is thrown up with a velocity of 78.4 m/south. Find how high it volition rise and how much fourth dimension it volition take to return to its point of project?

How to Calculate Distance from Acceleration and Velocity

Solution

Let $O$ be the bespeak of projection as shown in the effigy.

It is given in the question that at time $t=0$

Initial velocity $u=78.4m/due south$ and

$x_0=0$

Let $t$ exist the time taken to attain the maximum summit. It must be noted that velocity at maximum superlative is zero. So,

Final velocity $v=0$

According to the question we have to notice the distance $x_t$ at time $t$ and time $2t$ which is the time it volition have to render to its indicate of projection.

Here in solving this question, we will use initial conditions and required equations of motion to find variables of motion asked in the question.

The torso thrown upward moves under the effect of the gravitational strength of Earth. So, hither nosotros will use standard gravity,

$grand = -9.8 1000/southward^two$ ,

for equations involving the Globe's gravitational forcefulness as the acceleration charge per unit of the object.

Here we take taken acceleration due to gravity as negative considering we have taken upward direction every bit positive.

Now,

$5^2-u^ii=2g[x_t-x_0]$

putting in all the values we get

$0^2-(78.four)^ii=two(-9.8)[x_t-0]$

Therefore,

$x_t=\frac{78.4\times 78.4}{2\times ix.8}m=313.6\,\, m$

Again from the first equation of motion which relates final velocity, initial velocity, acceleration, and time we get

$v=u+at$

plugging in required values we become

$0=7.4+(-9.viii)t$

or,

$9.8t=78.4$

0r,

$t=\frac{78.4}{9.8}s=8s$

Since the time of rising is equal to the time of descent. And so total time is taken,

$2t=16s$

Instance 2

The reaction time for an automobile driver is 0.7 seconds. If the machine can decelerate at $5m/due south^ii$, calculate the total distance traveled in coming to stop from an initial velocity of $30 Km/h$ after a signal is observed.

Note:-
☛ After reading this question you can annotation that nosotros are given deceleration and the initial velocity, and we have to detect the distance traveled in coming to finish.
☛ Here we are also aware that the terminal velocity of the motorcar is nothing considering the vehicle is coming to stop after the application of the restriction.
☛ Here in this question, nosotros accept to find the distance in two parts
1. Distance traveled during the response fourth dimension of 0.seven seconds.
two. Distance traveled while the machine decelerates.

Solution

Since the reaction time of the driver is $0.4s$ therefore the automobile, during this time, volition continue to move with uniform velocity of $xxx Km/h$. Since all other units are given in metric organization nosotros will convert $30 Km/h$ to $grand/s$. So,

$30 Km/h=30\times\frac{1000}{3600} k/s=\frac{25}{iii}m/southward$

Distance covered during 0.7 second

$=\frac{25}{3}\times 0.7m=v.83m$

Let us choose this time as reference fourth dimension $(t=0)$ when the auto begins to decelerate.

So, at

$t=0$,
$x_0=0$,
initial velocity is $u=\frac{25}{3}m/s$ and
acceleration $a=-5m/s^2$.

We have to observe the full distance traveled in coming to stop from initial velocity.

Putting these values in third equation of motion $v^ii=u^2+2as$ we accept

$0^2-\left ( \frac{25}{3} \correct )^2=2\times(-5)[x_t-0]$

or,

$x_t=\frac{625}{9}\times \frac{1}{ten}m=6.94m$

Full distance travelled $= 5.83 m+six.94 m=12.77 m$

Related Links

kinematics – How to get distance when acceleration is non abiding? – Physics Stack Exchange – An interesting read that discussed how to notice altitude when acceleration is not constant. Information technology involves calculations using calculus.
http://hyperphysics.phy-astr.gsu.edu/hbase/index.html – Concept maps of about all concepts of physics.
Units of measurement in metric system

Source: https://www.physicsgoeasy.com/how-to-calculate-distance-from-acceleration-and-velocity/

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