Question

two cars are stopped next to each other and at $t = 0$ the cars begin to move in the same direction. one car has a constant acceleration $a_1$ until time $t_1$, and the displacement of this car from 0 to $t_1$ is $d$. if the second car has a constant acceleration of $\frac{a_1}{2}$ until time $2t_1$, what is the displacement of the second car from 0 to $2t_1$ in terms of $d$?

a $\frac{d}{2}$
b $d$
c $2d$
d $4d$

Explanation:

Step1: Find displacement formula for first car

The displacement formula for an object with constant - acceleration starting from rest ($v_0 = 0$) is $x=v_0t+\frac{1}{2}at^{2}$. For the first car with initial velocity $v_0 = 0$, acceleration $a = a_1$ and time $t=t_1$, the displacement $D=\frac{1}{2}a_1t_1^{2}$.

Step2: Find displacement formula for second car

For the second car, the initial velocity $v_0 = 0$, acceleration $a=\frac{a_1}{2}$ and time $t = 2t_1$. Using the displacement formula $x=v_0t+\frac{1}{2}at^{2}$, we substitute the values: $x=\frac{1}{2}\times\frac{a_1}{2}\times(2t_1)^{2}$.

Step3: Simplify the displacement of the second car

First, expand $(2t_1)^{2}=4t_1^{2}$. Then $x=\frac{1}{2}\times\frac{a_1}{2}\times4t_1^{2}$. Simplify the right - hand side: $x=\frac{1}{2}a_1t_1^{2}\times2$. Since $D = \frac{1}{2}a_1t_1^{2}$, then $x = 2D$.

Answer:

C. $2D$