Kinematics

Question

A particle P moves in a straight line so that its displacement, s m, from a fixed point O , t seconds after passing through point A one the line , is given by s =t^2-3t^2-9t+30. Calculate 

 

(a) the speed P when t=2,

(b) the acceleration of P when t=4

(c) the distance of P from A when it is instantaneously at rest,

(d) the average speed over the first 4 seconds.

Answer 1

To find the velocity of the particle, you'll want to find the equation's derivative:

 

s = t³ - 3t² - 9t + 30

s' = 3t² - 6t - 9

 

Now, plug in t = 2 to find velocity s':

 

s' = 3(2)² - 6(2) - 9

s' = 3(4) - 12 - 9

s' = -9 m/s

 

That means the speed is 9 m/s.

 

 

 

To find the acceleration of the particle, you'll want to find the equation's second derivative:

 

s' = 3t² - 6t - 9

s'' = 6t - 6

 

Now, plug in t = 4 to find acceleration s'':

 

s'' = 6(4) - 6

s'' = 24 - 6

s'' = 18 m/s²

 

 

 

When the particle is "instantaneously at rest," its velocity is 0 m/s. So, plug in velocity s' = 0 and solve for t:

 

s' = 3t² - 6t - 9

0 = 3t² - 6t - 9

0 = t² - 2t - 3

0 = (t - 3)(t + 1)

t - 3 = 0    OR    t + 1 = 0

t = 3    OR    t = -1

 

We'll ignore t = -1 since (typically) time can't be negative. So, t = 3 seconds after the particle passes through point A, it is at rest. It is s = (3)³ - 3(3)² - 9(3) + 30 = 27 - 27 - 27 + 30 = 3 meters from point O.

 

You can graph the function to find that this is the particle's minimum distance from point O. Or, you can use the first derivative to find the function's critical points t = 3 and t = -1 (the times when s' equals 0). We found earlier that when t = 2, the function's slope is -9, a negative. At t = 4, the function's slope is 3(4)² - 6(4) - 9 = 48 - 24 - 9 = 15, a positive. Since the function goes from negative to positive across t = 3, the function reaches its local minimum at t = 3.

 

Remember that the shortest distance between a line and a point is the line perpendicular to the line through that point. All of that means that if we draw a triangle with vertices at O, A, and the particle's location at t = 3, we'll have a right triangle.

 

The distance from O to the particle's location at t = 3 is s = 3 meters, as we calculated before. The distance from O to A is the particle's distance from O at t = 0: s = (0)³ - 3(0)² - 9(0) + 30 = 30 meters. So, our right triangle has a leg of length 3 and a hypotenuse of length 30. We'd like to find the distance from A to the particle's location at t = 3, the second leg. Call that x. Then:

 

a² + b² = c²

3² + x² = 30²

9 + x² = 900

x² = 891

x ≈ 29.85 meters

 

So, the distance from P to A when P is at rest is around 29.85 meters.

 

You can use a similar method to find the distance from P to A at t = -1, if you'd like.

 

 

 

To find the average speed as opposed to the instantaneous speed, we'll need to use the formula distance = speed × time. We know that time t = 4, so we just need to figure out the distance from A to P at t = 4 seconds.

 

We can draw another right triangle with vertices O, P at t = 3, and P at t = 4. The distance from O to P at t = 3 is 3, as we calculated before. The distance from O to P at t = 4 is s = (4)3 - 3(4)2 - 9(4) + 30 = 64 - 48 - 36 + 30 = 10 meters. So, our right triangle has one leg of length 3 and a hypotenuse of length 10. We'd like to find the other leg, x. So:

 

3² + x² = 10²

9 + x² = 100

x² = 91

x ≈ 9.54 meters

 

So, the distance the particle moves from t = 3 to t = 4 is 9.54 meters. Add that to the distance the particle moved from t = 0 to t = 3 to get a total distance of 29.85 + 9.54 = 39.39 meters.

 

Now, plug that into the distance = speed × time equation:

 

39.39 = speed × 4

9.85 = speed

 

The particle's average speed over the first four seconds is 9.85 m/s.