This problem involves calculating the final position of a particle undergoing constant acceleration. We use the standard kinematic equation for motion with constant acceleration.
The position ($S$) of a particle at time ($t$) is given by:
$S = S_0 + V_0 t + \frac{1}{2} a t^2$
Where:
We are given:
Substitute these values into the equation:
$S = 0 + (12\ m/s)(10\ s) + \frac{1}{2} (-2\ m/s^2)(10\ s)^2$
The final position of the particle after 10 seconds is 20 meters.
Consider the motion of a point on a circular trajectory. The acceleration in a linear motion (a) and the acceleration in angular motion (α), are related as : (Take r as the radius of circular trajectory)
A body of mass 10 kg moving with a velocity of 1 m/s is acted upon by a force of 50 N for two seconds. The final velocity will be:
A ball is dropped on a smooth horizontal surface from height ‘h’. What will be the height of rebounce after second impact, if coefficient of restitution between ball and surface is ‘e’?
Each of four particles move along an x-axis. Their coordinates (in meters) as functions of time (in seconds) are given by
1) particle 1: x (t) = 3.5 – 2.7 t3
2) particle 2: x (t) = 3.5 + 2.7 t3
3) particle 3: x (t) = 3.5 – 2.7 t2
4) particle 4: x (t) = 3.5 – 3.4t - 2.7 t2
Which of these particles have constant acceleration?
If water in a stream is flowing with a velocity of 20 kmph and a boat is travelling from one bank to another bank, if the velocity of boat in a direction perpendicular to direction of stream is 20 kmph and width of the stream is 2km, then the time taken and the angle at which boat makes with the direction stream is,