Given that a particular moving object\'s velocity is given by the equati on (t)

Question

Given that a particular moving object's velocity is given by the equati on (t) 65-3t, what is the equati on for the object's accelerati on'? A. a(t)-3 B. a(t)-3 C. a(t)-65 D. a(t) = t Suppose the position of an object on a number line is given by p(t) t2 3t +1 where t is in seconds and p is in centimeters. What is the acceleration of the object when t-2 seconds? A. 4 cm/sec2 B. 2 cm/sec C. 1 cm/s D. -1 cm/sec2 Given that a particular moving object's velocity is given by the equation v(e) -65t -22, what is the equation for the object's accelerati on? A, a(t) = 65 + 4t B. alt) 65 C. a(t) = 65-4t D. a(t) 61 Suppose an object falling out of an airplane has a velocity of v(t)--32t 64 where t is in seconds and vis in feet per second. What is the acceleration of the object when t 1? ft sec ft sec A. B. 32 ft sec ft sec C. 32 D. -962 Consider this graph of P'(t). At which t-value is P(t) the smalist? A. C. t3 E. Don't know 2

Explanation / Answer

(1)

Acceleration = v'(t)

So, a(t) = d/dt(65-3t) => a(t)=-3

So choice (B) is correct.

(2)

a(t) = p"(t)

So a(t) = 2

Hence acceleration at t=2 seconds is 2 cm/sec^2

So choice (B) is correct.

(3)

a(t) = v'(t)

=> a(t) = 65-4t

Hence choice (C) is correct .

(4)

a(t) = v'(t)

So a(t) = -32

Hence choice ( C ) is correct.

(5)

when derivative changes sign from negative to positive, then it is point of maxima.

So t=t3 is point of maxima, and there is no point in the graph. at which derivative changes sign from negative to positive.

Hence choice (E) is correct.

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