Question

A sprinter has to cover a total run if 100m. She increases her speed from rest under a uniform acc. if 1.0 ms^-2 up to 3 quarters of the total run and covers the last quarter with uniform speed. Find the time she takes to cover the 1st half and to cover the 2nd half of the run.​

Answer 1

Answer:-

$\red{\bigstar}$ The time taken to cover the 1st half:-

$\large\leadsto\boxed{\sf{10 \: sec.}}$

$\red{\bigstar}$ The time taken to cover the 2nd half :-

$\large\leadsto\boxed{\sf{4.25 \: sec.}}$

• Given:-

Total distance = 100 m

Acceleration = 1.0 ms-²

• To Find:-

Time taken to cover the 1st half and to cover the 2nd half of the run = ?

• Solution:-

Let s = 0 and t = 0 be starting location and time. Let t1 , t2 and t3 be the time when the sprinter is at the distance 50 m, 75 m and 100 m . The initial velocity of the sprinter, u = 0 as starts from rest and the acceleration is a = 1.0 ms-² [given].

Using,

$\pink{\bigstar}$$\boxed{\sf{s = ut + \dfrac{1}{2}a t ^2}}$

as u = 0 (starting from rest) and t = 0 ( starting point and time is 0)

therefore,

$\boxed{\sf{s = \dfrac{1}{2}a t ^2}}$

• The time $\bf{t_1}$ taken to cover the 1st half [50 m]:-

→ $\sf{50 = \dfrac{1}{2} \times 1.0 \times t_{1}^{2}}$

→ $\sf{t_{1} = \sqrt{100}}$

→ 10 sec.

• The time $\bf{t_2}$ taken to cover 3 quarters [75 m]:-

→ $\sf{75 = \dfrac{1}{2} \times 1.0 \times t_{2}^{2}}$

→ $\sf{t_{2} = \sqrt{150}}$

→ 12.2 sec.

• Time taken to cover the distance from 50m to 75m :-

→ $\sf{t_{2} - t_{1}}$

→ 12.2 - 10

→ 2.2 sec.

• The velocity of the sprinter at 75m :-

We know,

$\pink{\bigstar}$$\boxed{\sf{v = u + at}}$

here, u = 0 and time[t] is $\bf{t_2}$

hence,

$\boxed{\sf{v = a t_{2}}}$

→ 1.0 × 12.2

→ 12.2 ms-¹

The sprinter covers the last quarter from 75m to 100m = 25m with uniform velocity of 12.2 ms-¹.

Time taken:-

$\sf{t_3 - t_2 = \dfrac{25 m}{12.2 ms^{-1}}}$

→ 2.05 sec.

Hence,

The time taken to cover the 2nd half [50m to 100m] will be:-

→ 2.2 + 2.05 sec.

→ 4.25 sec.

Answer 2

Answer:

Answer:-

\red{\bigstar}★ The time taken to cover the 1st half:-

\large\leadsto\boxed{\sf{10 \: sec.}}⇝

10sec.

\red{\bigstar}★ The time taken to cover the 2nd half :-

\large\leadsto\boxed{\sf{4.25 \: sec.}}⇝

4.25sec.

• Given:-

Total distance = 100 m

Acceleration = 1.0 ms-²

• To Find:-

Time taken to cover the 1st half and to cover the 2nd half of the run = ?

• Solution:-

Let s = 0 and t = 0 be starting location and time. Let t1 , t2 and t3 be the time when the sprinter is at the distance 50 m, 75 m and 100 m . The initial velocity of the sprinter, u = 0 as starts from rest and the acceleration is a = 1.0 ms-² [given].

Using,

\pink{\bigstar}★ \boxed{\sf{s = ut + \dfrac{1}{2}a t ^2}}

s=ut+

2

1

at

2

as u = 0 (starting from rest) and t = 0 ( starting point and time is 0)

therefore,

\boxed{\sf{s = \dfrac{1}{2}a t ^2}}

s=

2

1

at

2

• The time \bf{t_1}t

1

taken to cover the 1st half [50 m]:-

→ \sf{50 = \dfrac{1}{2} \times 1.0 \times t_{1}^{2}}50=

2

1

×1.0×t

1

2

→ \sf{t_{1} = \sqrt{100}}t

1

=

100

→ 10 sec.

• The time \bf{t_2}t

2

taken to cover 3 quarters [75 m]:-

→ \sf{75 = \dfrac{1}{2} \times 1.0 \times t_{2}^{2}}75=

2

1

×1.0×t

2

2

→ \sf{t_{2} = \sqrt{150}}t

2

=

150

→ 12.2 sec.

• Time taken to cover the distance from 50m to 75m :-

→ \sf{t_{2} - t_{1}}t

2

−t

1

→ 12.2 - 10

→ 2.2 sec.

• The velocity of the sprinter at 75m :-

We know,

\pink{\bigstar}★ \boxed{\sf{v = u + at}}

v=u+at

here, u = 0 and time[t] is \bf{t_2}t

2

hence,

\boxed{\sf{v = a t_{2}}}

v=at

2

→ 1.0 × 12.2

→ 12.2 ms-¹

The sprinter covers the last quarter from 75m to 100m = 25m with uniform velocity of 12.2 ms-¹.

Time taken:-

\sf{t_3 - t_2 = \dfrac{25 m}{12.2 ms^{-1}}}t

3

−t

2

=

12.2ms

−1

25m

→ 2.05 sec.

Hence,

The time taken to cover the 2nd half [50m to 100m] will be:-

→ 2.2 + 2.05 sec.

→ 4.25 sec.