Question

Two cars start off to race with velocities 5 m/s and 8 m/s travel in straight line with uniform accelerations 4 m/s2 and 1 m/s2 respectively. What is the length of the path, if they reach the final point at the same time?

Answer 1

Answer:

18 metres

Explanation:

Given:

• Initial velocity of first car = u = 5 m/s

• Initial velocity of second car = u = 8 m/s

• Acceleration of first car = a = 4 m/s²

• Acceleration of second car = a = 1 m/s²
• Distance and time are constant for both the cars

To find:

• Length of the path

Using second equation of motion for first car:

S=$5t+\frac{1}{2} \times 4t^{2}$

S=$5t+2t^{2}$

Using second equation of motion for second car:

S=$8t+\frac{1}{2} \times 1t^{2}$

S=$8t+ \frac{1}{2} t^{2}$

As the distance of both the cars are equal :

$5t+2t^{2}=8t+\frac{1}{2} t^{2}$

$8t-5t=2t^{2} -\frac{1}{2}t^{2}$

$3t=\frac{3}{2} t^{2}$

Solving the above we get t = 2 seconds

Now substituting the value of t:

S=$5 \times 2 +\frac{1}{2} \times 4 \times 2^{2}$

S=10+8

S= 18 metres

The length of the path equals to 18 metres

Answer 2

Given :-

Initial velocity of Car A = 5m/s

Acceleration of car A = 4m/s^2

Initial velocity of Car= 8m/s

Acceleration of Car =1m/s^2

Time and distance are equal for both cars (constant)

To Find :-

The length of path

By using 2nd equation of motion

$\boxed{\sf {s= ut+\dfrac{1}{2}at^2}}$

Putting values -

• Distance of Car A

$\longrightarrow\sf s= 5\times t +\dfrac{1}{\cancel2}\times \cancel{4}\times t^2\\ \\ \longrightarrow\sf s_1= 5t+2t^2-----(1)$

• Distance of car B

$\longrightarrow\sf s= 8\times t+\dfrac{1}{2}\times 1\times t^2\\ \\ \longrightarrow\sf s_2= 8t+\dfrac{t^2}{2}-----(2)$

• Now it is given that both car covers same distance At same time

$\sf s_1= s_2\\ \\ \longrightarrow\sf 5t+2t^2= 8t+\dfrac{t^2}{2}\\ \\ \longrightarrow\sf 2t^2-\dfrac{t^2}{2}= 8t-5t\\ \\ \longrightarrow\sf \dfrac{4t^2-t^2}{2}= 3t\\ \\ \longrightarrow\sf 3t^2= 3t\times 2\\ \\ \longrightarrow\sf \cancel{\dfrac{3t^2}{3t}}=2\\ \\ \longrightarrow\sf t= 2$

• Time taken by bus to cover the distance is. 2 seconds
• Now find the distance put the value of time in equation 2

$\longrightarrow\sf s_2= 8t+\dfrac{t^2}{2}\\ \\ \longrightarrow\sf s_2= 8\times 2+\dfrac{(2)^2}{2}\\ \\\longrightarrow\sf s_2= 16+\cancel{\dfrac{4}{2}}\\ \\ \longrightarrow\sf s_2= 16+2 = 18m$

$\boxed{\sf\ \ length \ of \ path = 18m}$