A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?

Here are the solutions to the questions. 3. A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how? Yes, it is possible for her scooter to be accelerating even if the speedometer reading is constant. Acceleration is the rate of change of velocity. Velocity is a vector quantity, meaning it has both magnitude (speed) and direction. A speedometer measures only speed (the magnitude of velocity). If the scooter is moving at a constant speed but changing its direction, its velocity is changing, and therefore it is accelerating. This occurs when the scooter is moving in a circular path or any curved path. The acceleration in this case is called centripetal acceleration. 4. A car starts from rest and its velocity reaches 24 m s^-1 in 6 s. Find the average acceleration and the distance travelled in these 6 s. Step 1: Identify the given values. Initial velocity, u = 0 m s^-1 (starts from rest) Final velocity, v = 24 m s^-1 Time, t = 6 s Step 2: Calculate the average acceleration. The formula for average acceleration is: a = (v - u)/(t) Substitute the given values: a = 24 m s^-1 - 0 m s^-16 s a = 24 m s^-16 s a = 4 m s^-2 Step 3: Calculate the distance travelled. The formula for distance travelled with constant acceleration is: s = ((u + v)/(2))t Substitute the given values: s = (0 m s^-1 + 24 m s^-12) × 6 s s = (24 m s^-12) × 6 s s = 12 m s^-1 × 6 s s = 72 m The average acceleration is 4 m s^-2. The distance travelled is 72 m. 5. A motorbike moving with initial velocity 28 m s^-1 and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop. Step 1: Identify the given values. Initial velocity, u = 28 m s^-1 Final velocity, v = 0 m s^-1 (stops) Distance, s = 98 m Step 2: Calculate the acceleration. Using the kinematic equation v^2 = u^2 + 2as: (0 m s^-1)^2 = (28 m s^-1)^2 + 2 × a × 98 m 0 = 784 m^2 s^-2 + 196a m Rearrange to solve for a: -196a m = 784 m^2 s^-2 a = 784 m^2 s^-2-196 m a = -4 m s^-2 The negative sign indicates deceleration. Step 3: Calculate the time taken to come to a stop. Using the kinematic equation v = u + at: 0 m s^-1 = 28 m s^-1 + (-4 m s^-2) × t Rearrange to solve for t: -28 m s^-1 = -4 m s^-2 × t t = -28 m s^-1-4 m s^-2 t = 7 s The acceleration of the motorbike is -4 m s^-2. The time taken to come to a stop is 7 s. 3 done, 2 left today. You're making progress.