Question

A car is driven from 0-s if the car travelling initially with 36 km -h-1 is stopped by the driver after observing a signal by deceleration of 5 m-s-2 the total distance travelled by the car before coming to rest is
A) 18 m
B) 14m
C) 12m
D) 10m

Answer 1

Answer:

Provided that:

• Initial velocity = 36 kmph
• Final velocity = 0 mps
• Acceleration = -5 mps sq.

Don't be confused!

→ Final velocity cames as zero because it is given that the driver stopped the car.

→ Acceleration cames in negative because it is given that the car is retarding or it decelerates. And we also know that deceleration is the inverse of acceleration.

To calculate:

• The distance

Solution:

• The distance = 10 m

Using concepts:

• Formula to convert kmph-mps.
• Third equation of motion.

Using formulas:

• ${\small{\underline{\boxed{\pmb{\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}}}}}$

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

Required solution:

~ Firstly let us convert initial velocity into metre per second!

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

• Final velocity = 0 mps
• Initial velocity = 10 mps

~ Now let's calculate the distance by using third equation of motion!

$:\implies \sf v^2 \: - u^2 \: = 2as \\ \\ :\implies \sf (0)^{2} - (10)^{2} = 2(-5)(s) \\ \\ :\implies \sf 0 - 100 = -10s \\ \\ :\implies \sf -100 = -10s \\ \\ :\implies \sf 100 = 10s \\ \\ :\implies \sf \dfrac{100}{10} \: = s \\ \\ :\implies \sf 10 \: = s \\ \\ :\implies \sf s \: = 10 \: m$