Question

A car is moving on a frictionless (in the radial direction) circular banked road, with
a banking angle of $15^{\circ}$, find the velocity of the car.
Mass of car is $600 \mathrm{kg}$, the radius of the circular road $=10 \mathrm{m}$
Assume $\mathrm{g}=9.81 \mathrm{m} / \mathrm{s}^{2}$
(A) $4.26 \mathrm{m} / \mathrm{s}$
(B) $5.12 \mathrm{m} / \mathrm{s}$
(C) $51.2 \mathrm{m} / \mathrm{s}$
(D) $42.6 \mathrm{m} / \mathrm{s}$

Answer 1

Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.

Complete step by step answer
We know that momentum is a physics term that refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion (on the move) then it has momentum. One example is the use of air bags in automobiles. Air bags are used in automobile.s because they are able to minimize the effect of the force on an object involved in a collision. Air bags accomplish this by extending the time required to stop the momentum of the driver and passenger. Momentum is mass in motion, and any moving object can have momentum. An object's change in momentum is equal to its impulse. Impulse is a quantity of force times the time interval. Impulse is not equal to momentum itself; rather, it's the increase or decrease of an object's momentum. Based on this concept we have to solve this question

We know that the force acting on the car = $\operatorname{mgtan}(\theta)$ $\Rightarrow \dfrac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{mg} \tan (\theta)$
$\Rightarrow \mathrm{v}^{2}=\mathrm{rg} \tan (\theta)$
$\Rightarrow \mathrm{v}^{2}=10 \times 9.81 \times 0.268$
$\Rightarrow \mathrm{v}=5.12 \mathrm{m} / \mathrm{s}$

Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.