Q28E - University Physics with Modern Physics

Question

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 m/s (45 mi/) when it reaches the end of the 120-mlong ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Step-by-Step Solution

Answer

(a) Acceleration of the car= 1.66 m/s²

(b) Time taken by car to travel the length of ramp = 12.04 s.

(c) The distance travelled = 240.8 m.

Step 1: Identification of the given data

Speed of the car (v) =20 m/s

Length of the ramp (s) =120 m 

Initial velocity (u) = 0

Step 2: calculation for the acceleration

(a) 

 

Acceleration of the car can be calculated by using,

${\mathrm{v}}^2={\mathrm{u}}^2+(2\times \mathrm{a}\times \mathrm{s})$ 

Where, v= final velocity, u=initial velocity, a=acceleration due to gravity, and s=distance traveled

$$\begin{gathered} {\mathrm{v}}^2={\mathrm{u}}^2+(2\times \mathrm{a}\times \mathrm{s}) \\ {20}^2=0+(2\times \mathrm{a}\times 120) \\ \mathrm{a}=1.66\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the value of acceleration is $\mathrm{a}=1.66\mathrm{m}/{\mathrm{s}}^2$

Step 3: calculation for the time taken

(b) 

 

As we know that,

$\mathrm{v}=\mathrm{a}\times \mathrm{t}$ 

Then the required for car to travel ramp

$$\begin{gathered} t=\frac{20}{1.66} \\ t=12.04s \end{gathered}$$ 

Thus, the time taken by the car t = 12.04 s 

Step 4: calculation for the distance

(c)

$d=v\times t$ Substituting the values,

$$\begin{gathered} d=20 m/s\times 12.04s \\ d=240.8m \end{gathered}$$ 

Thus, the distance covered by the car d=24.08 m