
The velocity of the train increases uniformly from $20Kmh{r^{ - 1}}$ to $60Kmh{r^{ - 1}}$ in $4hrs$. Find the distance travelled during this period is:
Answer
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Hint:To solve this question, we need to know the three equations of motions. First of all, we need to calculate the value of acceleration. After that we have to select the equation which will give the value of distance travelled and we just have to substitute the required values to get the distance.
Formulae used:
$v = u + at$
Here $v$ is the final velocity of the train,
$u$ is the initial velocity of the train,
$a$ is the acceleration of the train and
$t$ is the time required for the journey by the train.
$S = ut + \dfrac{1}{2}a{t^2}$
Here $S$ is the distance travelled by the train in the given time,
$u$ is the initial velocity of the train,
$a$ is the acceleration of the train and
$t$ is the time required for the journey by the train.
Complete step by step answer:
In the question,
$u = 20Kmh{r^{ - 1}}$
$v = 60Kmh{r^{ - 1}}$
$t = 4hrs$
Where $v$ is the final velocity of the train,
$u$ is the initial velocity of the train and
$t$ is the time required for the journey by the train.
To solve this question, we have to first calculate the acceleration of the train. We can use one of the equations of motion to find acceleration.
We know that,
$ \Rightarrow v = u + at$
Here $v$ is the final velocity of the train,
$u$ is the initial velocity of the train,
$a$ is the acceleration of the train and
$t$ is the time required for the journey by the train.
Substituting the values of initial velocity, final velocity and time, the acceleration will be
$ \Rightarrow a = \dfrac{{v - u}}{t}$
$ \Rightarrow a = \dfrac{{60 - 20}}{4} = 10Kmh{r^{ - 2}}$
Now we can use another equation of motion to calculate distance.
We know that,
$ \Rightarrow S = ut + \dfrac{1}{2}a{t^2}$
Here $S$ is the distance travelled by the train in the given time,
$u$ is the initial velocity of the train,
$a$ is the acceleration of the train and
$t$ is the time required for the journey by the train.
Substituting the values of the initial velocity, time and acceleration we get,
\[ \Rightarrow S = \left( {20 \times 4} \right) + \left( {\dfrac{1}{2} \times 10 \times {4^2}} \right) = 160Km\]
So the distance travelled will be $160Km$.
Note:While using the equations of motion, be careful while substituting the values of initial and final velocities. The sign of acceleration should be taken correctly as sometimes retardation can also occur for which we need to use a negative sign. Also, the units are very important. All the values should be in the same system as the unit. If they are not in the same system then convert them all in the same system.
Formulae used:
$v = u + at$
Here $v$ is the final velocity of the train,
$u$ is the initial velocity of the train,
$a$ is the acceleration of the train and
$t$ is the time required for the journey by the train.
$S = ut + \dfrac{1}{2}a{t^2}$
Here $S$ is the distance travelled by the train in the given time,
$u$ is the initial velocity of the train,
$a$ is the acceleration of the train and
$t$ is the time required for the journey by the train.
Complete step by step answer:
In the question,
$u = 20Kmh{r^{ - 1}}$
$v = 60Kmh{r^{ - 1}}$
$t = 4hrs$
Where $v$ is the final velocity of the train,
$u$ is the initial velocity of the train and
$t$ is the time required for the journey by the train.
To solve this question, we have to first calculate the acceleration of the train. We can use one of the equations of motion to find acceleration.
We know that,
$ \Rightarrow v = u + at$
Here $v$ is the final velocity of the train,
$u$ is the initial velocity of the train,
$a$ is the acceleration of the train and
$t$ is the time required for the journey by the train.
Substituting the values of initial velocity, final velocity and time, the acceleration will be
$ \Rightarrow a = \dfrac{{v - u}}{t}$
$ \Rightarrow a = \dfrac{{60 - 20}}{4} = 10Kmh{r^{ - 2}}$
Now we can use another equation of motion to calculate distance.
We know that,
$ \Rightarrow S = ut + \dfrac{1}{2}a{t^2}$
Here $S$ is the distance travelled by the train in the given time,
$u$ is the initial velocity of the train,
$a$ is the acceleration of the train and
$t$ is the time required for the journey by the train.
Substituting the values of the initial velocity, time and acceleration we get,
\[ \Rightarrow S = \left( {20 \times 4} \right) + \left( {\dfrac{1}{2} \times 10 \times {4^2}} \right) = 160Km\]
So the distance travelled will be $160Km$.
Note:While using the equations of motion, be careful while substituting the values of initial and final velocities. The sign of acceleration should be taken correctly as sometimes retardation can also occur for which we need to use a negative sign. Also, the units are very important. All the values should be in the same system as the unit. If they are not in the same system then convert them all in the same system.
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