Example Question - physics formula
Here are examples of questions we've helped users solve.
Calculation of Force from Mass and Acceleration
F = m \cdot a \\ F = 5 \text{ kg} \cdot 3 \text{ m/s}^2 \\ F = 15 \text{ N}
Physics Problem: Stopping a Car on a Wet Road
The image contains a physics problem written in Italian. It describes a car with good pneumatics on a wet road that is able to brake with a constant deceleration of 4.92 m/s². There are two questions (Va) and (Vb) being asked based on this scenario: (Va) How much time does it take to stop when the initial velocity is 24.6 m/s? (Vb) How much space is needed to stop? Let's solve each question one at a time. For (Va): Time to stop (t) We can use the formula for deceleration (a) to find the time (t) it takes for the car to come to a stop with an initial velocity (u): v = u + at Since the final velocity (v) is 0 when the car stops, and we have the values for u and a, we can rearrange the formula to solve for t: 0 = 24.6 m/s + (-4.92 m/s²)t -24.6 m/s = -4.92 m/s²t t = 24.6 m/s / 4.92 m/s² t = 5 seconds For (Vb): Space needed to stop (d) We can use the formula that relates initial velocity, final velocity, deceleration, and distance: v² = u² + 2ad Again, the final velocity (v) is 0, and we have the other values: 0 = (24.6 m/s)² + 2*(-4.92 m/s²)d 0 = 605.16 m²/s² - 9.84 m/s²*d 9.84 m/s²*d = 605.16 m²/s² d = 605.16 m²/s² / 9.84 m/s² d = 61.5 meters So, the answers are: (Va) The time it takes for the car to come to a stop is 5 seconds. (Vb) The space needed for the car to stop is 61.5 meters.
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