Question

show all of your work.

1. a car, with a starting velocity of 10m/s accelerates to 35m/s over a period of 10s. using only a velocity/time graph (do not use formulas from the formula sheet), calculate the displacement of the car over the time interval. (4)

\\( \overrightarrow { v _ { i } } = 10 m / s \\)
\\( \overrightarrow { a } = 35 m / s \\)
\\( \delta t = 10 s \\)

1. a car is travelling with a velocity of 20m/s. the driver increases the cars speed at a rate of 0.5m/s² until the cars velocity is 25m/s. for how long, in seconds, was the car accelerating to reach the new speed of 25m/s? (3)

equation
\\( \delta \overrightarrow { d } = \left( \frac { \overrightarrow { v _ { f } } + \overrightarrow { v _ { i } } } { 2 } \
ight) \delta t \\)
\\( \overrightarrow { v _ { f } } = \overrightarrow { v _ { i } } + \overrightarrow { a } _ { a v } \delta t \\)
\\( \delta \overrightarrow { d } = \overrightarrow { v _ { i } } \delta t + \frac { 1 } { 2 } \overrightarrow { a } _ { a v } \delta t ^ { 2 } \\)
\\( v _ { f } ^ { 2 } = v _ { i } ^ { 2 } + 2 a _ { a v } \delta d \\)
\\( \delta \overrightarrow { d } = \overrightarrow { v _ { f } } \delta t - \frac { 1 } { 2 } \overrightarrow { a } _ { a v } \delta t ^ { 2 } \\)

Explanation:

Step1: Identify the formula

We use the formula \(v_{f}=v_{i}+a_{av}\Delta t\). Here, \(v_{i} = 20m/s\), \(v_{f}=25m/s\), \(a_{av}=0.5m/s^{2}\), and we need to find \(\Delta t\).

Step2: Rearrange the formula

From \(v_{f}=v_{i}+a_{av}\Delta t\), we can solve for \(\Delta t\) as \(\Delta t=\frac{v_{f}-v_{i}}{a_{av}}\).

Step3: Substitute the values

Substitute \(v_{i} = 20m/s\), \(v_{f}=25m/s\), \(a_{av}=0.5m/s^{2}\) into \(\Delta t=\frac{v_{f}-v_{i}}{a_{av}}\). So \(\Delta t=\frac{25 - 20}{0.5}\).

Step4: Calculate the result

\(\Delta t=\frac{5}{0.5}=10s\).

Answer:

\(10s\)