How do I find displacement using a velocity time graph using the following question as an example with a step by step procedure?
The displacement in a velocity-time graph is the total area above the x-axis minus the area below the x-axis till the time mentioned in the question. Although this is fairly straightforward, this may explain things more clearly - Displacement from time and velocity example :)
The displacement of a particle is given by x=(t-2) ^2. What is the distance covered by the particle in the first 4 seconds?
Although you can refer the answer given by Mr Kulkarni but if you want to go logically then just try to imagine the graph of x versus time of the displacement of this object initially the object was 4 unit away from origin but after 2 second it comes to origin and then again after 2 seconds it gets back to its original coordinates hence the total distance travelled is equals to 4 + 4 that is 8 units but if you want to know that how will you do a complicated version of this problem that is in which the XT function is not such simple then either as I told you should refer Mr Kulkarni’s method and you also have one more approach you just need to find the region of the curve where the velocity is negative that is slope of the XT curve is negative which is in between the region of zero to 2 seconds in the given problem you just break the problem in two regions from initial stage to the stage where the slope becomes zero and then again from the time wear 2.0 to the final time if the slope becomes zero more time in between the duration then you can repeat this again and again and break the problem in as many parts as slope become zero in the duration.Hope it helped.
Velocity (distance and displacement) calculus?
The velocity function is -t^2+4t-3 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [0,5] For the Total Displacement I got (-20/3) which is correct. I am unable to find the distance to this, could someone provide me with a formula and answer for the distance?? When I did try and follow a formula I got -20/3 again for the distance.. I tried -20/3 and 20/3 for the distance and both were incorrect.
Displacement of particle help?
To get the distance traveled in need to integrate v(t) from t=-2 to t=3. Integral( t^3 - 2t^2, -2 to 3) = [1/4^t4 - 2/3t^3]_-2 to 3 = { (1/4*(3^4)-(2/3 *3^3)- 1/4(-2)^4-2/3(-2)^3 } = (1/4 * 81 - 2/3 * 27 ) - (1/4 * 16 - 2/3 * (-8)) = (81/4 - 18) - ( 4 + 16/3) = 2.25 - 9.333 = -7.08m Displacement you have to determine the direction. Since it is along a line, it is either positive or negative. Displacement is -7.08 m, e.g. displaced 7.08m to the left ( negative x-direction or the to the left) Distance traveled = absolute value of displacement = | -7.08m| = 7.08 m The velocity at t=-2 is v(-2) = (-2)^3 - 2 * (-2)^2 = -6 - 2*4= -14 m/s, so the particle is moving to the left (negative) direction initially. v(3) = (3)^3 - 2 * (3)^2 = 27 - 2*9 = 27 - 18 = 9 m/s, so at the end the velocity is positive and the particle moving to the right. Some time in the middle the velocity changed (note v(t=0)=0). So from t=-2 to t=0, the change in distance is: [1/4^t4 - 2/3t^3]_-2 to 0 = s(t=0) - s(t-2) = = 1/4*0-2/3*0 - 1/4 * 16 - 2/3 * (-8)) = 0 - (4+16/3) = 28/3= -9 1/3 s(t=3)-s(t=0) = (81/4 - 18) - 0 = +2.25 so it move to the left more than it has move to the right from it's original starting point at time t=-2.
Calc 1 distance traveled?
v(t) = t² - 2t - 15 Let s(t) be the displacement at time t, where s(0) = 0 is an arbitrary initial condition. s(t) = ∫ (t² - 2t - 15) dt = t³/3 - t² - 15t + C s(0) = 0 (0)³/3 - (0)² - 15(0) + C = 0 C = 0 s(t) = t³/3 - t² - 15t s(6) - s(1) = (6)³/3 - (6)² - 15(6) - [(1)³/3 - (1)² - 15(1)] = -54 - (-47/3) = -115/3 From time 1 to time 6, the displacement is -115/3 m. Let v(t) = 0. t² - 2t - 15 = 0 (t + 3)(t - 5) = 0 t = -3 or t = 5 At time 1, the velocity is negative. The particle moves backwards until it comes to a rest at time 5. After that, the velocity is positive, moving the particle forward until time 6. s(5) = (5)³/3 - (5)² - 15(5) = -175/3 Distance traveled from time 1 to time 6 = |s(5) - s(1)| + |s(6) - s(5)| = |-175/3 - (-47/3)| + |-54 - (-175/3)| = 128/3 + 13/3 = 47 m