Question

A car accelerates uniformly in a straight line with acceleration of 10m/s2 and travels 150m in a time interval of 5s how far would he travel in the next 5s

Answer 1

Answer:

Given:

Acceleration of car = 10 m/s

Distance travelled in 5 sec = 150 m

To find:

Distance travelled in the next 5 seconds

Concept:

There are 2 ways to app this kind of questions .

Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.

Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.

Calculation:

v² = u² + 2as

=> ( u + at)² = u² + 2as

=> u² + 2uat + a²t² = u² + 2as

=> 2uat + (at)² = 2as

=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)

=> 100u = 3000 - 2500

=> 100u = 500

=> u = 5 m/s

Distance travelled in 10 seconds :

s = ut + ½at²

=> s = (5 × 10) + ½(10)(10)²

=> s = 50 + 500

=> s = 550 m

Distance travelled in the 2nd half will be :

d = 550 - 150

=> d = 400 m

So final answer is :

$\boxed{ \green{ \huge{ \bold{d = 400 \: m}}}}$

Answer 2

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

• Distance covered in next 5s is 400 m

$\rule{200}{0.5}$

$\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}$

Given :

• Acceleration (a) = 10 m/s²
• Distance traveled in 5s = 150 m

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To Find :

• Distance traveled in next 5 sec

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Solution :

• First we are finding the initial velocity

Use 2nd equation of motion :

$\large{\boxed{\sf{s \: = \: ut \: + \: \dfrac{1}{2} at^2}}} \\ \\ \implies {\sf{150 \: = \: u \: \times \: 5 \: + \: \dfrac{1}{2} \: \times \: 10 \: \times \: 5^2}} \\ \\ \implies {\sf{150 \: = \: 5u \: + \: 5 \: \times \: 25}} \\ \\ \implies {\sf{150 \: = \: 5u \: + \: 125}} \\ \\ \implies {\sf{5u \: = \: 150 \: - \: 125}} \\ \\ \implies {\sf{5u \: = \: 25}} \\ \\ \implies {\sf{u \: = \: \dfrac{25}{5}}} \\ \\ \implies {\sf{u \: = \: 5}} \\ \\ \underline{\sf{\therefore \: Initial \: velocity \: is \: 5 \: ms^{-1}}}$

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• Now, distance traveled in 10 s

Again use 2nd equation of motion :

$\large{\boxed{\sf{s \: = \: ut \: + \: \dfrac{1}{2} at^2}}} \\ \\ \implies {\sf{s \: = \: 5 \: \times \: 10 \: + \: \dfrac{1}{2} \: \times \: 10 \: \times \: 10^2}} \\ \\ \implies {\sf{s \: = \: 50 \: + \: 5 \: \times \: 100}} \\ \\ \implies {\sf{s \: = \: 50 \: + \: 500}} \\ \\ \implies {\sf{s \: = \: 550}} \\ \\ \underline{\sf{\therefore \: Distance \: covered \: in \: 10 \: s \: is \: 550 \: m}}$

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So,

distance traveled in next 5s = distance in 10s - distance in first 2 s

⇒s = 550 - 150

⇒s = 400

So, distance traveled in next 5s is 400 m