The relationship between the displacement of a particle and time is: S = 4T + 2 times the square of T, the units of S and T are meter and second respectively, The relationship between displacement and time is: S = 4 * t + 2 * t * t Therefore, the relationship between velocity and time is v = DS / dt = 4 + 4 * t So the initial velocity of the particle is: V0 = 4 + 4 * 0 = 4m / s The relationship between acceleration and time is a = DV / dt = 4 So the acceleration of particle is a = 4m / S2 The relationship between the velocity and time is v = DS / dt = 4 + 4 * t Can you tell me how to push it out? We haven't learned derivative yet Please tell me v=ds/dt=4+4*t My feeling is that v = DS / dt = 4 + 2 * t is right

The relationship between the displacement of a particle and time is: S = 4T + 2 times the square of T, the units of S and T are meter and second respectively, The relationship between displacement and time is: S = 4 * t + 2 * t * t Therefore, the relationship between velocity and time is v = DS / dt = 4 + 4 * t So the initial velocity of the particle is: V0 = 4 + 4 * 0 = 4m / s The relationship between acceleration and time is a = DV / dt = 4 So the acceleration of particle is a = 4m / S2 The relationship between the velocity and time is v = DS / dt = 4 + 4 * t Can you tell me how to push it out? We haven't learned derivative yet Please tell me v=ds/dt=4+4*t My feeling is that v = DS / dt = 4 + 2 * t is right

You've learned a formula! S = VT + 1 / 2at square! You see, 4T in the formula means V: 4 times time t, and the same is true after that! Do you understand? Do you understand the above? That's the direct formula! Kinematics formula: average velocity v = displacement divided by time velocity v = initial velocity V + acceleration a times time t, how can