Question

1 a car accelerates from rest to 6.6 m/s in 6.5 seconds. find the distance the car traveled. record your answer and fill in the bubbles on your answer document. be sure to use the correct place value. 2 a car traveling at 15 m/s in a straight line accelerates for 5s with an acceleration of 2 m/s2. what is its final velocity? record your answer and fill in the bubbles on your answer document. be sure to use the correct place value. 3 a car is traveling at 15 m/s. it goes down a hill where it accelerates at 3 m/s² for 4s. how fast is it going at the bottom of the hill? record your answer and fill in the bubbles on your answer document. be sure to use the correct place value. 4 pete will be attempting a long jump. starting from rest, he wants to reach a top speed of 9 m/s and he can accelerate at 4 m/s². how many meters must his approach be? record your answer and fill in the bubbles on your answer document. be sure to use the correct place value.

Explanation:

Step1: Identify the relevant kinematic - equation

For the first problem, since the car starts from rest ($u = 0$), we use the equation $s=ut+\frac{1}{2}at^{2}$. First, we find the acceleration $a=\frac{v - u}{t}$. Here, $u = 0$, $v = 6.6$ m/s and $t = 6.5$ s. So, $a=\frac{6.6-0}{6.5}=\frac{6.6}{6.5}$ m/s². Then, using $s=ut+\frac{1}{2}at^{2}$ with $u = 0$, we have $s=\frac{1}{2}\times\frac{6.6}{6.5}\times(6.5)^{2}$.

Step2: Calculate the distance for the first problem

$s=\frac{1}{2}\times6.6\times6.5 = 21.45$ m.

Step3: For the second problem

We use the equation $v=u + at$. Given $u = 15$ m/s, $a = 2$ m/s² and $t = 5$ s. Then $v=15+2\times5$.

Step4: Calculate the final velocity for the second problem

$v=15 + 10=25$ m/s.

Step5: For the third problem

We use the equation $v=u+at$. Given $u = 15$ m/s, $a = 3$ m/s² and $t = 4$ s. Then $v=15+3\times4$.

Step6: Calculate the final velocity for the third problem

$v=15 + 12=27$ m/s.

Step7: For the fourth problem

Since Pete starts from rest ($u = 0$), and we want to find the distance $s$ with a final velocity $v = 9$ m/s and acceleration $a = 4$ m/s². We use the equation $v^{2}-u^{2}=2as$. Since $u = 0$, we have $s=\frac{v^{2}}{2a}$.

Step8: Calculate the distance for the fourth problem

$s=\frac{9^{2}}{2\times4}=\frac{81}{8}=10.125$ m.

Answer:

1. $21.45$ m
2. $25$ m/s
3. $27$ m/s
4. $10.125$ m