Two cars $P$ and $Q$ start side by side from rest at the same instant and in the same direction.$P$ accelerates uniformly at $4\,m/{s^2}$ for $10$ seconds and then moves with uniform velocity.$Q$ moves with uniform acceleration $3\,m/{s^2}$. Find the distance between them $30$ Seconds after start.
Answer
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Hint: We know the value of the uniform speed of acceleration here. Acceleration is determined by dividing velocity by time, then dividing the meter by seconds per second. Dividing distance twice by time is often the same as dividing the distance by the square of time. At the starting point, we get the interval between seconds.
Formula used:
Acceleration formula,
$\overline a = \dfrac{{v - {v_0}}}{t} = \dfrac{{\Delta v}}{{\Delta t}}$
Where,
$\overline a $ is average acceleration
$v$ is final velocity
${v_0}$ is starting velocity
$t$ is elapsed time
Complete step by step solution:
Given by,
$P$ accelerates uniformly at $4\,m/{s^2}$ for $10$ seconds
$Q$ moves with uniform acceleration $3\,m/{s^2}$
We find the after $30s$ starting distance,
Now,
Movement of $P$
First $10$ seconds accelerated motion with acceleration $4\,m/{s^2}$,
According to the acceleration formula,
Speed one ${S_1} = \dfrac{1}{2} \times 4 \times 10 \times 10$
On simplifying,
We get,
$\Rightarrow$ ${S_1} = 200m$
Speed after $10$ seconds velocity,
$\Rightarrow$ $v = 4 \times 10$
On simplifying,
We get,
$\Rightarrow$ $40\,m/s$
Distance travelled next $20$seconds,
$\Rightarrow$ ${S_2} = 40 \times 20$
Here,
$\Rightarrow$ ${S_2} = 800\,m/s$
Then,
Total distance travelled in $30$ seconds
$\Rightarrow$ ${S_1} + {S_{_2}}$ substituting the given value,
$\Rightarrow$ $200 + 800 = 1000\,m$
Now,
Movement of $Q$
Distance travelled in $30$ seconds
$\Rightarrow$ $S = \dfrac{1}{2} \times 3 \times 30 \times 30$
On simplifying,
We get,
$\Rightarrow$ $S = 1350\,m$
Here we subtract the both movement of $P$ and $Q$
$\Rightarrow$ $1000 - 1350$
We get,
$\Rightarrow$ $350\,m$
Hence $Q$ is ahead by $350\,m$.
Thus the $350\,m$ distance after start the $30$ seconds.
Note: When acceleration often happens if there is a non-zero net force acting on an object, the acceleration value is negative here because the car slows decelerating. Universal gravitation which states that with an exponential force, any two objects with mass will attract each other.
Formula used:
Acceleration formula,
$\overline a = \dfrac{{v - {v_0}}}{t} = \dfrac{{\Delta v}}{{\Delta t}}$
Where,
$\overline a $ is average acceleration
$v$ is final velocity
${v_0}$ is starting velocity
$t$ is elapsed time
Complete step by step solution:
Given by,
$P$ accelerates uniformly at $4\,m/{s^2}$ for $10$ seconds
$Q$ moves with uniform acceleration $3\,m/{s^2}$
We find the after $30s$ starting distance,
Now,
Movement of $P$
First $10$ seconds accelerated motion with acceleration $4\,m/{s^2}$,
According to the acceleration formula,
Speed one ${S_1} = \dfrac{1}{2} \times 4 \times 10 \times 10$
On simplifying,
We get,
$\Rightarrow$ ${S_1} = 200m$
Speed after $10$ seconds velocity,
$\Rightarrow$ $v = 4 \times 10$
On simplifying,
We get,
$\Rightarrow$ $40\,m/s$
Distance travelled next $20$seconds,
$\Rightarrow$ ${S_2} = 40 \times 20$
Here,
$\Rightarrow$ ${S_2} = 800\,m/s$
Then,
Total distance travelled in $30$ seconds
$\Rightarrow$ ${S_1} + {S_{_2}}$ substituting the given value,
$\Rightarrow$ $200 + 800 = 1000\,m$
Now,
Movement of $Q$
Distance travelled in $30$ seconds
$\Rightarrow$ $S = \dfrac{1}{2} \times 3 \times 30 \times 30$
On simplifying,
We get,
$\Rightarrow$ $S = 1350\,m$
Here we subtract the both movement of $P$ and $Q$
$\Rightarrow$ $1000 - 1350$
We get,
$\Rightarrow$ $350\,m$
Hence $Q$ is ahead by $350\,m$.
Thus the $350\,m$ distance after start the $30$ seconds.
Note: When acceleration often happens if there is a non-zero net force acting on an object, the acceleration value is negative here because the car slows decelerating. Universal gravitation which states that with an exponential force, any two objects with mass will attract each other.
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