Question

Ben Rushin is waiting at a stoplight in his car. When the light turns green, Ben accelerates from rest at a rate of 6.00 m/s/s for an interval of 4.10 seconds. Determine the displacement of Ben’s car during this time period.

Answer 1

The displacement of Ben's car during this time period is 50.43 m.

Since Ben's accelerate his car uniformly, therefore we can apply the equation of motion

The displacement of the Ben's car can be calculate by the 2nd equation of motion

S=ut+0.5a$t^{2}$

Here, u=0 initial velocity of the car, and S is displacement of car

Plugging the values in the above equation

S=0*4.1+0.5*6*4.1^2=50.43 m

Therefore, the displacement of Ben's car during this time period is 50.43 m.

Answer 2

Question :

Ben Rushin is waiting at a stoplight in his car. When the light turns green, Ben accelerates from rest at a rate of 6.00 m/s/s for an interval of 4.10 seconds. Determine the displacement of Ben’s car during this time period.

Answer :

Formula Used :-

Second Equation Of Motion Formula :

$\begin{gathered}\mapsto \sf\boxed{\bold{\orange{s =\: ut + \dfrac{1}{2} at^2}}}\\\end{gathered}$

where,

s = Distance Covered or displacement

u = Initial Velocity

t = Time

a = Acceleration

Now,

$\begin{gathered}\bigstar\: \: \rm{\bold{Initial\: Velocity\: (u) =\: 0\: m/s}}\\\end{gathered}$

$\begin{gathered}\bigstar\: \: \rm{\bold{Time\: (t) =\: 4.10\: seconds}}\\\end{gathered}$

$\begin{gathered}\bigstar\: \: \rm{\bold{Acceleration\: (a) =\: 6\: m/s^2}}\\\end{gathered}$

According to the question by using the formula we get,

$\begin{gathered}\longrightarrow \sf s =\: 0 \times 4.10 + \dfrac{1}{2} \times 6 \times (4.10)^2\\\end{gathered}$

$\begin{gathered}\longrightarrow \sf s =\: 0 + \dfrac{1}{2} \times 6 \times 4.10 \times 4.10\\\end{gathered}$

$\begin{gathered}\longrightarrow \sf s =\: 0 + \dfrac{\cancel{6}}{\cancel{2}} \times \bigg(\dfrac{410}{100}\bigg) \times \bigg(\dfrac{410}{100}\bigg)\\\end{gathered}$

$\begin{gathered}\longrightarrow \sf s =\: 0 + 3 \times \bigg(\dfrac{1681\cancel{00}}{100\cancel{00}}\bigg)\\\end{gathered}$

$\begin{gathered}\longrightarrow \sf s =\: 3 \times \bigg(\dfrac{1681}{100}\bigg)\\\end{gathered}$

$\longrightarrow \sf s =\: 3 \times \dfrac{1681}{100}$

$\longrightarrow \sf s =\: \dfrac{5043}{100}$

$\longrightarrow \sf\bold{\purple{s =\: 50.43\: m}}$

∴ The displacement of Ben's car during this time period is 50.43 m