To solve this problem, we can use the equations of motion. We are given that a car is moving with uniform acceleration and passes through two points, P and Q, with velocities of 30 km/h and 40 km/h, respectively. We want to find the velocity of the car at the midpoint between P and Q.
Let's break down the problem step by step:
1. Given data:
- Initial velocity at point P (u) = 30 km/h
- Final velocity at point Q (v) = 40 km/h
- Distance between P and Q (s)
2. We can use the following equation of motion to relate velocity, initial velocity, acceleration, and distance:
\[v^2 = u^2 + 2as\]3. We want to find the velocity of the car (V) at the midpoint between P and Q. Let's call this point M.
4. First, let's find the acceleration (a) of the car using the data for points P and Q:
\[40^2 = 30^2 + 2a s\]Solving for 'a':
\[a = \frac{40^2 - 30^2}{2s} = \frac{1600 - 900}{2s} = \frac{700}{2s} = \frac{350}{s}\]5. Now, we can find the velocity at point M using the same equation of motion:
\[V^2 = u^2 + 2a \left(\frac{s}{2}\right)\]\[V^2 = (30^2) + 2 \left(\frac{350}{s}\right) \left(\frac{s}{2}\right)\]\[V^2 = 900 + 350\]\[V^2 = 1250\]Taking the square root of both sides:
\[V = \sqrt{1250} \approx 35.35 \text{ km/h}\]So, the velocity of the car at the midpoint between points P and Q is approximately 35.35 km/h.
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is:
The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion.
Linear motion is also known as the Rectilinear Motion which are of two types: