Question

Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.

Answer 1

Given:-

• Initial velocity ,u = 0m/s
• Acceleration ,a = 6m/s
• Time taken ,t = 4.10s

To Find:-

• Displacement ,s

Solution:-

According to the Question

It is given that Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds.

We have to calculate the displacement of the Ben's car .

Using 2nd equation of motion

• s = ut + 1/2at²

where,

• s denote displacement
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

→ s = 0×4.1 + 1/2×6 × (4.10)²

→ s = 0 + 3 × 16.81

→ s = 3×16.81

→ s = 50.43m

• Hence, the displacement of the Ben's car is 50.43 metres.

Answer 2

Answer:

Given :-

• Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6 m/s² for a time of 4.10 seconds.

To Find :-

• What is the displacement of Ben's car during this time period.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2} at^2}}}$

where,

• s = Distance Covered or displacement
• u = Initial Velocity
• t = Time
• a = Acceleration

Solution :-

Given :

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

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

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

According to the question by using the formula we get,

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

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

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

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

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

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

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

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

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