2. A particle moves along a straight line so that its acceleration at time t seconds is (t + 1)^{2} cm/sec^{2}. The particle’s position at time t = 0 is at the origin, and its initial velocity is 1 cm/sec. What is the position of the particle, in cm. at time t seconds?
A. ((t+1)^{4}/12)+(2/3)t-1/12 |
B. ((t+1)^{4}/12)+(2/3)t+1/12 |
C. ((t+1)^{4}+2t-1)/3 |
D. ((t+1)^{4}+2t+1)/3 |
E. ((t+1)^{4}-1)/4 |
Solution: The answer is A
Let,
a(t) = acceleration of the particle at any time t (seconds)
v(t) = velocity of the particle at any time t (seconds)
s(t) = the position of the particle at any time t (seconds)
Since a(t) is (t + 1)^{2} cm/sec^{2} and is also defined as the rate of change of v(t) with respect to time, we have the following differential equation,
Multiplying both sides of the above equation by dt and then integrating both sides yields the following equation,
where c_{1} is the constant of integration. Since the initial velocity is 1 cm/sec, that is v(0)=1, we can solve for c_{1} by substituting this result in the above equation, as follows,
Thus,
Since v(t) is also defined as the rate of change of s(t) with respect to time, we have the following differential equation,
Multiplying both sides by dt and integrating both sides yields,
where c_{2} is the constant of integration. Since the particle’s position at t=0 is the origin, that is, s(0)=0, then clearly by substitution in the above equation c_{2}=0. So, the position of the particle, in cm, at time t seconds is given by,