Q34E - University Physics with Modern Physics

Question

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 2.80 m/s². At the same instant a truck, travelling with a constant speed of 20.0 m/s, overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck? (b) How fast is the car travelling when it overtakes the truck? (c) Sketch an x-t graph of the motion of both vehicles. Take x = 0 at the intersection. (d) Sketch a v_x-t graph of the motion of both vehicles.

Step-by-Step Solution

Answer

(a) The total distance = 308.3 m.

(b) Speed of the car = 41.59 m/s.

Step 1: Identification of the given data

• Constant acceleration (a) = 2.80 m/s²
• Truck constant speed (v) = 20.8 m/s

Step 2: calculation for the distance

(a) 

 

The distance is calculated by using,

For the truck $d=v\times t$  ………………….. (1)

And also for the car $d=v_0+\left(\frac{1}{2}\times a\times t^2\right)$ ………….. (2)

Comparing equations (1) and (2) we get,

 

$v\times t=v_0\times t+\left(\frac{1}{2}\times a\times t^2\right)$………………….. (3)

Putting the given values in equation (3),

$$\begin{gathered} 20.8\times t=0+\left(\frac{1}{2}\times 2.80\times t^2\right) \\ t=\frac{2\times 20.8\mathrm{m}/\mathrm{s}}{2.80\mathrm{m}/{\mathrm{s}}^2} \\ t=14.85\mathrm{s} \end{gathered}$$ 

Then the distance will be,

$$\begin{gathered} \mathrm{d}=\mathrm{v}\times \mathrm{t} \\ \mathrm{d}=20.8 \mathrm{m}/\mathrm{s}\times 14.85\mathrm{s} \\ "\mathrm{d}=308.8" \mathrm{m} \end{gathered}$$ 

Thus, the total distance will be d =308.8 m

Step 3: calculation for the speed of the car

(b)

 

Velocity is calculated by using,

$v=a\times t$

Substituting the value

$$\begin{gathered} \mathrm{v}=2.80\mathrm{m}/{\mathrm{s}}^2\times 14.85\mathrm{s} \\ \mathrm{v}=41.59 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

Thus, the speed of the car v=41.59 m/s .

Step 4: x-t graph

(c)

Step 5: Vx-t graph

(d)