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Question

A particle moves along a line so that at any time t, where t is greater than or equal to zero and less than or equal to pi, its position is given by

s(t) = -4cost - t^2/2 + 10

What is the velocity of the particle when its acceleration is zero.
PLease please please show how you get the answer.

You're the best.

Answer 1

the previous answer in not true. If acceleration is zero, that means velocity is a constant, not zero. The calculus answer to this question is:
V(t) = 4sint - t
A(t) = 4cost - 1
setting a = 0 gives 4cost = 1, t = arccos(0.25) = 1.318
plugging t into V(t) formula gives
V(1.318) = 4sin(1.318) - 1.318 = 2.55

Answer 2

You have an expression for position as a function of time. The first derivative of that function is velocity and the 2'nd derivative is acceleration. So set the 2'nd derivative equal to zero, solve for t, and substitute the t value back into the 1'st derivative to get velocity (at time t). You can also substitute that value of t into the original expression to find the -position- of the particle when the acceleration is zero ☺


Doug

Answer 3

If acceleration is zero, then velocity is zero

Answer 4

v(t)=s'(t)=4sin(t)-t
a(t)=s''(t)=4cos(t)-1

When acceleration is 0, cos(t)=1/4. Then t=1.32.
At t=1.32, v(t)=4sin(1.32)-(1.32)=2.56.

Acceleration=0 does not mean velocity=0.

Answer 5

when a=0 then velocity is constant.
differciate s(t) u'l get the value.


u r welcome.