Question

A car decelerates from 10m/s2 to 5 m/s 2 in 2 seconds. What is the distance travelled by the car? a. 10 m b. 15 m c. 2 m d. 5 m

Answer 1

Proper question: A car retards from 10 m/s to 5 m/s in 2 seconds. What is the distance travelled by the car?

Provided that:

• Initial velocity = 10 m/s
• Final velocity = 5 m/s
• Time = 2 seconds

To calculate:

• The distance

Solution:

• The distance = 15 m

Using concepts:

• To calculate the acceleration we can use acceleration formula or first equation of motion too.

• Choice may vary!

• To calculate the distance we can use either second equation of motion and third equation of motion.

• Choice may vary!

Using formulas:

• Acceleration formula,

• ${\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}$

• First equation of motion,

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

• Second equation of motion,

• ${\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}}$

• Third equation of motion,

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

Where, a denotes acceleration, u denotes initial velocity, t denotes time taken, v denotes final velocity, s denotes displacement or distance or height.

Required solution:

~ Firstly let us find acceleration!

Finding accordingly acceleration formula!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{5-10}{2} \\ \\ :\implies \sf a \: = \dfrac{-5}{2} \\ \\ :\implies \sf a \: = -2.5 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -2.5 \: ms^{-2}$

Finding accordingly first equation of motion!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 5 = 10 + a(2) \\ \\ :\implies \sf 5 = 10 + (2a) \\ \\ :\implies \sf 5 - 10 = 2a \\ \\ :\implies \sf -5 = 2a \\ \\ :\implies \sf a \: = \dfrac{-5}{2} \\ \\ :\implies \sf a \: = -2.5 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -2.5 \: ms^{-2}$

~ Now let's find out the distance!

Finding accordingly third equation of motion!

$:\implies \sf 2as \: = v^2 \: - u^2 \\ \\ :\implies \sf 2(-2.5)(s) = (5)^{2} - (10)^{2} \\ \\ :\implies \sf 2(-2.5)(s) = 25 - 100 \\ \\ :\implies \sf -5s = -75 \\ \\ :\implies \sf 5s = 75 \\ \\ :\implies \sf s \: = \dfrac{75}{5} \\ \\ :\implies \sf s \: = 15 \: m \\ \\ :\implies \sf Distance \: = 15 \: m$

Now finding accordingly second equation of motion!

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 10(2) + \dfrac{1}{2} \times -2.5(2)^{2} \\ \\ :\implies \sf s \: = 10(2) + \dfrac{1}{2} \times -2.5 \times 4 \\ \\ :\implies \sf s \: = 20 + \dfrac{1}{2} \times -2.5 \times 4 \\ \\ :\implies \sf s \: = 20 + 1 \times -2.5 \times 2 \\ \\ :\implies \sf s \: = 20 + 1 \times -5 \\ \\ :\implies \sf s \: = 20 + (-5) \\ \\ :\implies \sf s \: = 20 - 5 \\ \\ :\implies \sf s \: = 15 \: m \\ \\ :\implies \sf Distance \: = 15 \: m$

• Henceforth, 15 metres is the distance travelled, therefore, option b) is absolutely correct!