Question

A toy car starts from rest with an acceleration of 2m/s^2. Two seconds later another identical toy car is released from the same position with the same acceleration.What is the distance between the two cars one second after the release of the second car?

Answer 1

Answer

The distance between the two cars one second after the release of the second car is 8m

$\bf\large\underline{Given}$

• A toy car starts from rest with an acceleration of 2m/s².
• Two seconds after the release of first car another identical car is released from same position with same acceleration.

$\bf\large\underline{To \ Find}$

• The distance between the two cars one second after the release of the second car.

$\bf\large\underline{Solution}$

For the first car

initial velocity , u = 0m/s

acceleration , a = 2m/s²

∵ the other car is released after 2s and we have to calculate after 1s of the release of second car, thus

time for the second car ,

t = 3s

From equations of motion ,

$\sf\rightarrow s_{1} = ut + \dfrac{1}{2} at^{2} \\\\ \sf\implies s_{1}= 0\times 3 + \dfrac{1}{2} \times 2 \times 3^{2} \\\\ \sf\implies s_{1} = 9m$

For the second car ,

initial velocity , u = 0m/s

acceleration , a = 2m/s²

the car is released after 2s of first and it covered distance for 1s so ,

t = 1s

From equations of motion

$\sf s_{2} = 0\times1 + \dfrac{1}{2}\times2 \times 1^{2} \\\\ \sf\implies s_{2} = 1m$

Therefore , distance between them in that particular second is

➜ s₁ - s₂ = 9m - 1m

➜ s₁ - s₂ = 8m