Question

an electron accelerated (with acceleration $a_1$) from rest for a time interval $t_1$ and travels a distance of 10 m.
a second electron given the acceleration $a_1/4$ for a time interval 4$t_1$, after starting from rest, will travel a distance of:
(a) 160 m
(b) 0.4 m
(c) 2.5 m
(d) 10 m
no, try using proportional reasoning, acceleration and time come with different exponents.
(e) 40 m

Explanation:

Step1: Recall the kinematic - equation

The kinematic equation for an object starting from rest ($v_0 = 0$) is $x=\frac{1}{2}at^{2}$, where $x$ is the distance traveled, $a$ is the acceleration, and $t$ is the time - interval. For the first electron, $x_1=\frac{1}{2}a_1t_1^{2}$, and we know $x_1 = 10$ m.

Step2: Apply the kinematic equation to the second electron

For the second electron, $a_2=\frac{a_1}{4}$ and $t_2 = 4t_1$. Substitute these values into the kinematic equation $x_2=\frac{1}{2}a_2t_2^{2}$. Then $x_2=\frac{1}{2}(\frac{a_1}{4})(4t_1)^{2}$.

Step3: Simplify the expression for $x_2$

Expand the right - hand side of the equation: $x_2=\frac{1}{2}\times\frac{a_1}{4}\times16t_1^{2}$. Since $\frac{1}{2}\times\frac{1}{4}\times16 = 2$, and $\frac{1}{2}a_1t_1^{2}=x_1 = 10$ m, we have $x_2 = 2\times(\frac{1}{2}a_1t_1^{2})$.

Step4: Calculate the distance $x_2$

Substitute $x_1 = 10$ m into the equation for $x_2$. So $x_2=2\times10=20$ m. But there is a mistake above. Let's start from step 2 again.
$x_2=\frac{1}{2}(\frac{a_1}{4})(4t_1)^{2}=\frac{1}{2}\times\frac{a_1}{4}\times16t_1^{2}=\frac{1}{2}a_1t_1^{2}\times4$. Since $\frac{1}{2}a_1t_1^{2}=10$ m, then $x_2 = 40$ m.

Answer:

E. 40 m