A particle starts from origin at t = 0 with a velocity
5i m/s and moves in xy-plane under action of a force
which produces a constant acceleration of (3i +2j) m/s?.
(i) What is the y-coordinate of the particle at the instant
its x-coordinate is 84 m?
(ii) What is the speed of the particle at this time?
[NCERT]
Answers
x(t) = Vx0*t + 0.5*Ax*t^2
y(t) = Vy0*t + 0.5*Ay*t^2
Vx0 = initial horizontal velocity
Vy0 = initial vertical velocity
Ax = horizontal acceleration
Ay = vertical acceleration
Based on the problem statement, we can easily see that Ax = 3 m/s^2 and Ay = 2 m/s^2, Vx0 = 5 m/s, and Vy0 = 0. So we can rewrite the above 2 equations as:
x(t) = 5.0*t + 1.5*t^2
y(t) = 1.0*t^2
Now we need to calculate the corresponding time, t, for x = 84 m:
x(t) = 5.0*t + 1.5*t^2
84 = 5.0*t + 1.5*t^2
t = 6.00 or -9.33
The t = -9.33 seconds is non-physical, so we must use t = 6.00 seconds. At t = 6.00, y(t) becomes:
y(t) = 1.0*t^2
y(t=6.00) = 1.0*6.00^2
y(t=6.00) = 36 m
Therefore, when the x coordinate of the particle is 84 m, the y coordinate will be 36 m. Hope that helps!