
A bus starts from rest and moves with constant acceleration $8m{s^{ - 2}}$. At the same time, a car travelling with constant velocity $16m{s^{ - 1}}$ overtakes and passes the bus. After how much time and at what distance, the bus overtakes the car?
A. $t = 4s,d = 64m$
B. $t = 5s,d = 72m$
C. $t = 8s,d = 58m$
D. None of the above
Answer
233.1k+ views
Hint First using Newton's third equation of motion derive an expression of distance where the bus overtakes the car. Then calculate the distance travelled by the car at the same time. Equate these two equations to obtain the values.
Formula used
$S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the distance travelled, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.
Complete step by step answer
This problem can be solved by implementing the equations of motions.
These equations of motions describe the behavior of a physical system based on their initial velocity, final velocity, acceleration and time.
Motion can be classified into two basic types- dynamics and kinematics.
In dynamics the forces and energies of the particles are taken into account. Whereas in kinematics only the position and time of the particle are taken into consideration.
The three main equations of motions are
$v = u + at$
${v^2} = {u^2} + 2aS$
$S = ut + \dfrac{1}{2}a{t^2}$
Where $v$ is the final velocity, $u$ is the initial velocity, $t$ is the time taken, $a$ is the acceleration and $S$ is the displacement of the body.
Let $t$ be the time and $d$ be the distance at which the bus overtakes the car.
Since the bus starts from rest, so its initial velocity must be zero i.e. $u = 0$
Using the third equation of motion,
$S = ut + \dfrac{1}{2}a{t^2}$ we get,
$d = \dfrac{1}{2} \times 8 \times {t^2}$
For the car, the distance is given as
$
d = vt \\
\Rightarrow d = 16t \\
$
Substituting the value of the distance in the equation of motion we get,
$
16t = \dfrac{1}{2} \times 8{t^2} \\
\Rightarrow t = 4s \\
$
Therefore, $d = 16 \times 4 = 64m$
So, the bus takes $4s$ to overtake the car at a distance of $64m$
Hence, the correct answer is option A.
Note The equations of motion give us a comprehensive idea about the behavior of a body in motion. They are based on the three Newton’s laws of motion.
Formula used
$S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the distance travelled, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.
Complete step by step answer
This problem can be solved by implementing the equations of motions.
These equations of motions describe the behavior of a physical system based on their initial velocity, final velocity, acceleration and time.
Motion can be classified into two basic types- dynamics and kinematics.
In dynamics the forces and energies of the particles are taken into account. Whereas in kinematics only the position and time of the particle are taken into consideration.
The three main equations of motions are
$v = u + at$
${v^2} = {u^2} + 2aS$
$S = ut + \dfrac{1}{2}a{t^2}$
Where $v$ is the final velocity, $u$ is the initial velocity, $t$ is the time taken, $a$ is the acceleration and $S$ is the displacement of the body.
Let $t$ be the time and $d$ be the distance at which the bus overtakes the car.
Since the bus starts from rest, so its initial velocity must be zero i.e. $u = 0$
Using the third equation of motion,
$S = ut + \dfrac{1}{2}a{t^2}$ we get,
$d = \dfrac{1}{2} \times 8 \times {t^2}$
For the car, the distance is given as
$
d = vt \\
\Rightarrow d = 16t \\
$
Substituting the value of the distance in the equation of motion we get,
$
16t = \dfrac{1}{2} \times 8{t^2} \\
\Rightarrow t = 4s \\
$
Therefore, $d = 16 \times 4 = 64m$
So, the bus takes $4s$ to overtake the car at a distance of $64m$
Hence, the correct answer is option A.
Note The equations of motion give us a comprehensive idea about the behavior of a body in motion. They are based on the three Newton’s laws of motion.
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