Question

A body started from rest and moved with uniform acceleration of 2 −1 for 10 seconds. Calculate its final velocity and distance travelled.​

Answer 1

Explanation:

The initial velocity travelled is 625m

The accleration is α=10m/s

2

The time is t=5s

Apply the equation of motion

v=u+at

The velocity is v=0+10.5=50m/s.

The distance travelled in the first 5s is

s=ut+

2

1

at

2

=0.5+

2

1

.10.5

2

=125m

Thedistancetravelledinthefollowing10sis

s

1

=ut

1

=50.10=500m

Thetotaldistancetravelledis

d=s+s

1

=125+500=625m

Answer 2

Provided that:

• Initial velocity = 0 m/s
• Acceleration = 2 m/s²
• Time = 10 seconds

To calculate:

• Final velocity
• Distance travelled

Solution:

• Final velocity = 20 m/s
• Distance travelled = 100 m

Using concepts:

• By using first equation of motion, we have to calculate the final velocity!

• By using either third or second equation, we have to calculate the distance travelled!

• Choice may vary!

Using formulas:

• 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,

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

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

Required solution:

~ By using first equation of motion let us find out the final velocity!

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf v \: = 0 + 2(10) \\ \\ :\implies \sf v \: = 0 + 20 \\ \\ :\implies \sf v \: = 20 \: ms^{-1} \\ \\ :\implies \sf Final \: velocity \: = 20 \: ms^{-1}$

~ Now let's calculate the distance travelled by using either third equation of motion or second equation of motion!

By using third equation of motion!

$:\implies \sf v^2 \: - u^2 \: = 2as \\ \\ :\implies \sf (20)^{2} - (0)^{2} = 2(2)(s) \\ \\ :\implies \sf 400 - 0 = 4s \\ \\ :\implies \sf 400 = 4s \\ \\ :\implies \sf \dfrac{400}{4} \: = s \\ \\ :\implies \sf 100 \: = s \\ \\ :\implies \sf s \: = 100 \: m \\ \\ :\implies \sf Distance \: = 100 \: m$

By using second equation of motion!

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 0(10) + \dfrac{1}{2} \times 2 \times (10)^{2} \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} \times 2 \times (10)^{2} \\ \\ :\implies \sf s \: = 0 + 1 \times 100 \\ \\ :\implies \sf s \: = 0 + 100 \\ \\ :\implies \sf s \: = 100 \: m \\ \\ :\implies \sf Distance \: = 100 \: m$