Question

a car is moving at 300 m/s; brakes are applied and it stops after travelling 150m. Find the time taken to stop?​

Answer 1

Answer:

Provided that:

• Initial velocity = 300 m/s
• Final velocity = 0 m/s
• Distance = 150 m

To calculate:

• The time taken

Solution:

• The time taken = 1 second

Using concepts:

• Third equation of motion
• First equation of motion

Using formulas:

First equation of motion:

• ${\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at}}}}}}$

Third equation of motion:

• ${\small{\underline{\boxed{\pmb{\sf{v^2 \: - u^2 \: = 2as}}}}}}$

Explanation:

• As we are asked to calculate the time so before calculating the time, we have to find out the acceleration.

Required solution:

~ Firstly finding acceleration by using third equation of motion!

$:\implies \sf v^2 \: - u^2 \: = 2as \\ \\ :\implies \sf (0)^{2} - (300)^{2} = 2(a)(150) \\ \\ :\implies \sf 0 - 90000 = 300a \\ \\ :\implies \sf -90000 = 300a \\ \\ :\implies \sf \dfrac{-90000}{300} = a \\ \\ :\implies \sf \dfrac{-900}{3} = a \\ \\ :\implies \sf -300 \: = a \\ \\ :\implies \sf Acceleration = -300 \: mps^{-2} \\ \\ :\implies \sf Retardation = -300 \: mps^{-2}$

~ Now let's find out the time taken by using first equation of motion!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 300 + (-300)t \\ \\ :\implies \sf 0 = 300 + (-300t) \\ \\ :\implies \sf 0 = 300 - 300t \\ \\ :\implies \sf 0 - 300 = -300t \\ \\ :\implies \sf -300 = -300t \\ \\ :\implies \sf 300 = 300t \\ \\ :\implies \sf \dfrac{300}{300} = t \\ \\ :\implies \sf 1 = t \\ \\ :\implies \sf t \: = 1 \: seconds \\ \\ :\implies \sf Time \: = 1 \: seconds$