A particle starts from rest and moves with constant acceleration. If the particle travels 39 m in the seventh second, find the distance travelled in the tenth second. (b) A train of length 66.5 m is travelling with uniform acceleration $ \frac{4}{7} \text{ m s}^{-2} $. It meets another train of length 91 m travelling on a parallel track in the opposite direction with uniform acceleration $ \frac{8}{7} \text{ m s}^{-2} $. Their speeds at this moment are 18 m s$^{-1}$ and 24 m s$^{-1}$ respectively. (i) Find the time taken for the trains to pass each other. (ii) Find the distance between the trains 1 second later.

Photo AI

A particle starts from rest and moves with constant acceleration - Leaving Cert Applied Maths - Question 1 - 2015

Question 1

A particle starts from rest and moves with constant acceleration. If the particle travels 39 m in the seventh second, find the distance travelled in the tenth secon... show full transcript

Worked Solution & Example Answer:A particle starts from rest and moves with constant acceleration - Leaving Cert Applied Maths - Question 1 - 2015

Step 1

Find the distance travelled in the tenth second

96%

114 rated

Answer

To solve for the distance travelled in the tenth second, we first need to find the acceleration of the particle.

Using the equation for distance in a given second:

$s_n = ut + \frac{1}{2} a (2n - 1)$

Given that the particle travels 39 m in the seventh second:

$s_7 = \frac{1}{2} a (13) = 39 \implies a = 6 \text{ m s}^{-2}$

Now using this acceleration to find the distance travelled in the tenth second:

$s_{10} = \frac{1}{2} a (19) = \frac{1}{2} \times 6 \times 19 = 57 \text{ m}$

Step 2

Find the time taken for the trains to pass each other

99%

104 rated

Answer

Let ( t ) be the time taken for the trains to pass each other.

Using the speeds of the trains:

Train 1 speed: 18 m s $^{-1}$
Train 2 speed: 24 m s $^{-1}$

The relative distance to be covered is the sum of their lengths:

$s_1 + s_2 = 66.5 + 91 = 157.5 \text{ m}$

Setting up the equation:

$(18 + 24)t = 157.5 \implies 42t = 157.5 \implies t = \frac{157.5}{42} = 3.75 \text{ seconds}$

Step 3

Find the distance between the trains 1 second later

96%

101 rated

Answer

After 1 second, the position of each train will change according to their speeds and the time.

For Train 1:

$s_1 = 18 \times (3.75 + 1) = 18 \times 4.75 = 85.5 \text{ m}$

For Train 2:

$s_2 = 24 \times (3.75 + 1) = 24 \times 4.75 = 114 \text{ m}$

The total distance between the trains:

$d = (s_1 + 66.5 + s_2) - 91 = (85.5 + 66.5 + 114) - 91 = 175 \text{ m}$

A particle starts from rest and moves with constant acceleration - Leaving Cert Applied Maths - Question 1 - 2015

Worked Solution & Example Answer:A particle starts from rest and moves with constant acceleration - Leaving Cert Applied Maths - Question 1 - 2015

Find the distance travelled in the tenth second

Find the time taken for the trains to pass each other

Find the distance between the trains 1 second later

Join the Leaving Cert students using SimpleStudy...