The initial velocity of a car, vi, is 45 km/h in the positivex direction. The final velocity of the car, vf, is 66 km/h indirection that points 75 degrees above the positive x axis.&nbsp;a) sketch the vectors -vi, vf and change in v. (b) findthe magnitude and direction in change of velocity.&nbsp;For b..my answer was -37.93 km/h and -88.41 degrees for direction.

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The initial velocity of a car, vi, is 45 km/h in the positivex direction. The final velocity of the car, vf, is 66 km/h indirection that points 75 degrees above the positive x axis.&amp;nbsp;a) sketch the vectors -vi, vf and change in v. (b) findthe magnitude and direction in change of velocity.&amp;nbsp;For b..my answer was -37.93 km/h and -88.41 degrees for direction.

Aubree Mcintyre · Accepted Answer

The resultantt velocity is given by: Vres=vi2+vf2+2vi2vf2cos⁡θ=84.5558kmh arctan⁡[vfsin⁡θvi+vfcos⁡θ]=45.7799∘

RizerMix · Accepted Answer

Step 1:a) To sketch the vectors -vi, vf, and change in v, we can represent them as arrows on a coordinate system. Let&#039;s assume the positive x-axis is to the right, and the positive y-axis is upwards.- The vector -vi represents the initial velocity of the car. Since it is given as 45 km/h in the positive x-direction, we draw an arrow pointing to the right with a magnitude of 45 units.- The vector vf represents the final velocity of the car. It is given as 66 km/h in a direction that points 75 degrees above the positive x-axis. To represent this, we draw an arrow starting from the origin, making an angle of 75 degrees with the positive x-axis, and having a magnitude of 66 units.- The change in velocity, represented as Δv, can be obtained by subtracting the initial velocity vector from the final velocity vector. In this case, Δv = vf - (-vi), which simplifies to Δv = vf + vi. We draw an arrow starting from the tip of -vi and ending at the tip of vf.Step 2:b) To find the magnitude and direction of the change in velocity, we need to calculate the resultant vector from the tip of -vi to the tip of vf.Let&#039;s denote the magnitude of the change in velocity as |Δv| and the angle it makes with the positive x-axis as θ.Using the parallelogram law of vector addition, we can find |Δv|:|Δv|=(vf·cos(θ)+vi)2+(vf·sin(θ))2where θ is the angle between vf and the x-axis.To find the direction θ, we can use the following equation:tan(θ)=vf·sin(θ)vf·cos(θ)+viSolving these equations will give us the magnitude and direction of the change in velocity.Substituting the given values, we have:|Δv|=(66·cos(75∘)+45)2+(66·sin(75∘))2tan(θ)=66·sin(75∘)66·cos(75∘)+45Let&#039;s evaluate the expressions to find the magnitude and direction of the change in velocity.First, let&#039;s calculate the magnitude |Δv|:|Δv|=(66·cos(75∘)+45)2+(66·sin(75∘))2Calculating the value, we have:|Δv|≈(66·0.2588+45)2+(66·0.9659)2≈(17.1928+45)2+(63.7614)2≈62.19282+4053.1865≈3864.3971+4053.1865≈7917.5836≈88.98km/hSo, the magnitude of the change in velocity is approximately 88.98 km/h.Next, let&#039;s calculate the direction θ:tan(θ)=66·sin(75∘)66·cos(75∘)+45Calculating the value, we have:tan(θ)=66·0.965966·0.2588+45≈63.761417.1928+45≈63.761462.1928≈1.0251To find θ, we take the arctan of both sides:θ≈arctan(1.0251)≈46.74∘So, the direction of the change in velocity is approximately 46.74 degrees above the positive x-axis.Therefore, the magnitude of the change in velocity is approximately 88.98 km/h, and the direction is approximately 46.74 degrees above the positive x-axis.

user_27qwe · Accepted Answer

Answer:74.10 km/h−88.41∘Explanation:(a) Sketching the vectors:To sketch the vectors -𝐯i, 𝐯f, and Δ𝐯, we need to consider their magnitudes and directions.The initial velocity, 𝐯i, is given as 45 km/h in the positive x direction. We represent this vector as 𝐯i=45km/h𝐢^.The final velocity, 𝐯f, is given as 66 km/h in a direction that points 75 degrees above the positive x axis. To represent this vector, we first convert the magnitude and direction into vector components. The x component of 𝐯f is 66km/h·cos(75∘), and the y component is 66km/h·sin(75∘). Therefore, we have 𝐯f=(66km/h·cos(75∘))𝐢^+(66km/h·sin(75∘))𝐣^.The change in velocity, Δ𝐯, is calculated by subtracting the initial velocity from the final velocity: Δ𝐯=𝐯f−𝐯i.(b) Finding the magnitude and direction of the change in velocity:To find the magnitude of the change in velocity, we calculate the length of the vector Δ𝐯 using the Pythagorean theorem:|Δ𝐯|=(Δvx)2+(Δvy)2where Δvx and Δvy are the x and y components of Δ𝐯, respectively.To find the direction of the change in velocity, we use trigonometry to calculate the angle it makes with the positive x axis:θ=tan−1(ΔvyΔvx)where θ represents the angle.Now, let&#039;s calculate the magnitude and direction of the change in velocity.Δ𝐯=𝐯f−𝐯iΔ𝐯=[(66km/h·cos(75∘))−45km/h]𝐢^+(66km/h·sin(75∘))𝐣^Δ𝐯=[(66km/h·0.2588)−45km/h]𝐢^+(66km/h·0.9659)𝐣^To find the magnitude of Δ𝐯:|Δ𝐯|=(−37.93km/h)2+(63.71km/h)2|Δ𝐯|=1440.3049+4058.1041|Δ𝐯|=5498.409|Δ𝐯|≈74.10km/hTo find the direction of Δ𝐯:θ=tan−1(63.71km/h−37.93km/h)θ≈−88.41∘Therefore, the magnitude of the change in velocity is approximately 74.10 km/h, and the direction is approximately −88.41∘.

The initial velocity of a car, vi, is 45 km/h in the positivex direction. The final velocity of the car, vf, is 66 km/h indirection that points 75 deg

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