Question

Starting from rest at the origin, x0 = 0, a car accelerates in a straight line along the +x direction. For the time interval from t0 = 0 s to t1= 4 s, its velocity is given by v(t) = (−10 m/s2 )t + (6 m/s3)t2 . Then from t1 = 4 s to t2 = 10 s, it maintains a constant velocity v = 56 m/s in the same direction.
(i) Find the average acceleration for the first time interval, from t0 = 0 s to t1 = 4 s.
(ii) Find the average acceleration for the second time interval, from t1 = 4 s to t2 = 10 s
(iii) Find the instantaneous acceleration a(t) as a function of time for 0 s < t < 4 s.
(iv) Find the position x(t) as a function of time for 0 s < t < 4 s

Answer 1

Answer:

i) The average acceleration for the first time interval, from $t_{0} = 0$ to $t_{1} =4\ s$, is $14 \frac{m}{s^{2} }$ .

ii) The average acceleration for the second time interval, from $t_{1} =4\ s$ to  $t_{2} =10\ s$, is zero.

iii) The instantaneous acceleration as a function of time a(t)  between 0 s and 4 s is $a(t)=-10\frac{m}{s^{2} } + 12\frac{m}{s^{3} } t$ .

iv) The position as a function of time x(t) between 0 s and 4 s is $x(t)=-5\frac{m}{s^{2} } t^{2} + 2\frac{m}{s^{3} } t^{3}$ .

Explanation:

We are given as data that a car that is a rest, that means v₀=0, at the origin, x₀=0, accelerates in a straight line along the +x direction. We are told that from $t_{0} = 0$ to $t_{1} =4\ s$ the velocity is $v(t)=-10\frac{m}{s^{2} } t + 6 \frac{m}{s^{3} } t^{2}$ and that from $t_{1} =4\ s$ to  $t_{2} =10\ s$ it maintains a constant velocity of $v=56\frac{m}{s}$ in the same direction.

i) The average acceleration is simply the rate of change of velocity, it can be expressed as:

    $\bar{a}=\frac{\vartriangle v}{\vartriangle t} =\frac{v_{f}-v_{0} }{t_{f}-t_{0}}$

we have to replace the values in this equation, the only value we need to calculate is $v_{f}$

    $v(4 s)=-10\frac{m}{s^{2} } 4 s + 6 \frac{m}{s^{3} } (4 s)^{2}$

    $v(4 s)=-10\frac{m}{s^{2} } 4 s + 6 \frac{m}{s^{3} } 16 s^{2}$

    $v(4 s)= -40 \frac{m}{s} + 96 \frac{m}{s}= 56 \frac{m}{s}$

we get that  $v_{f}= 56 \frac{m}{s}$ .

Then we pur this value in the expression of the average acceleration

    $\bar{a}=\frac{\vartriangle v}{\vartriangle t} =\frac{v_{f}-v_{0}}{t_{f}-t_{0}}=\frac{56\frac{m}{s}-0\frac{m}{s} }{4 s - 0 s}$

the average acceleration for the first time interval is

                      $\bar{a}=14\frac{m}{s^{2} }$

ii) The average acceleration for the second time interval, from $t_{1} =4\ s$ to  $t_{2} =10\ s$, is zero because the velocity is constant. The average acceleration is the rate of change of velocity, if this magnitude remains constant then it follows that the acceleration is zero.

iii) We calculate the instantaneous acceleration, which is the acceleration at a specific moment in time, as the derivative of the velocity function.

Mathematically   $a(t)=\frac{d v(t)}{d t }$

so we get

                   $a(t)=-10 \frac{m}{s^{2} } + 12\frac{m}{s^{3} } t$

iv) To find the position in function of time x(t) we simply integrate the expression of v(t):

    $v(t)=\frac{d x(t)}{d t}= -10\frac{m}{s^{2} }t+ 6\frac{m}{s^{3} } t^{2}$

we get that

                  $x(t)=-5\frac{m}{s^{2} } t^{2} + 2 \frac{m}{s^{3} } t^{2}$