Question

iWhat is the distance covered by an object which accelerates from rest with a final velocity of 80 m/s if the time taken is 10 s?​

Answer 1

Provided that:

• Initial velocity = 0 m/s
• Final velocity = 80 m/s
• Time = 10 seconds

To calculate:

• Distance

Solution:

• Distance = 400 m

Required solution:

~ To solve this question firstly we have to calculate the acceleration of the object, we can find acceleration by using acceleration formula or by using first equation of motion too, so choice may vary!

Firstly let us find acceleration by using acceleration formula.

$:\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf {\small{\underline{\boxed{\pmb{\mathcal{A} \sf cceleration \: = \dfrac{Change \: in \: velocity}{Time} \: \bigstar}}}}} \\ \\ :\implies \sf Acceleration \: = \dfrac{80-0}{10} \\ \\ :\implies \sf Acceleration \: = \dfrac{80}{10} \\ \\ :\implies \sf Acceleration \: = 8 \: ms^{-2} \\ \\ \sf \bigstar \: {\underline{\boxed{\pmb{\mathcal{A} \sf cceleration \: = \frak{8 \: ms^{-2}}}}}}$

Now let us calculate acceleration by using first equation of motion.

$:\implies {\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at \: \: \bigstar}}}}}} \\ \\ :\implies \sf 80 = 0 + a(10) \\ \\ :\implies \sf 80 = 0 + a \times 10 \\ \\ :\implies \sf 80 = a \times 10 \\ \\ :\implies \sf \dfrac{80}{10} \: = a \\ \\ :\implies \sf 8 \: = a \\ \\ :\implies \sf Acceleration \: = 8 \: ms^{-2} \\ \\ \bigstar \: {\underline{\boxed{\pmb{\mathcal{A} \sf cceleration \: = \frak{8 \: ms^{-2}}}}}}$

~ Now let us calculate the distance, we can calculate it by using either third equation of motion or second equation of motion, choice may vary!

Using third equation of motion.

$:\implies {\small{\underline{\boxed{\pmb{\sf{v^2 \: - u^2 \: = 2as \: \: \bigstar}}}}}} \\ \\ :\implies \sf (80)^{2} - (0)^{2} = 2(8)(s) \\ \\ :\implies \sf 6400 - 0 = 2(8s) \\ \\ :\implies \sf 6400 = 16s \\ \\ :\implies \sf \dfrac{6400}{16} \: = s \\ \\ :\implies \sf 400 \: = s \\ \\ :\implies \sf s \: = 400 \: m \\ \\ \bigstar \: {\underline{\boxed{\pmb{\mathcal{D} \sf istance \: = \frak{400 \: m}}}}}$

Using second equation of motion.

$:\implies {\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2 \: \: \bigstar}}}}}} \\ \\ :\implies \sf s \: = 0(10) + \dfrac{1}{2} \times 8(10)^{2} \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 8(10)^{2} \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 8 \times 10 \times 10 \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 8 \times 100 \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 800 \\ \\ :\implies \sf s \: = 400 \: m \\ \\ \bigstar \: {\underline{\boxed{\pmb{\mathcal{D} \sf istance \: = \frak{400 \: m}}}}}$

Answer 2

Distance covered by the object 400 metres

Explanation :-

we have :-

→ Initial velocity (u) = 0 m/s

→ Final velocity (v) = 80 m/s

→ Time taken (t) = 10 sec

To Find,

• Distance covered

________________________

Firstly let's calculate acceleration (a) of the body

by using the first equation of Motion

v = u + at

• 80 = 0 + a(10)
• 80 = 10a
• a = 80/10
• a = 8 m/s²

Now, we can calculate the distance covered by the object either by using the 2nd or 3rd equation of Motion

v² - u² = 2as

• (80)² - 0 = 2(8)(s)
• 6400 = 16s
• s = 6400/16
• s = 400m