Question

Starting from rest on a level road a girl can reach the speed of 5 m/s in 10s on a bicycle. Find

a. the acceleration.
b. the average speed during the 10s.
b. the distance she travelled in 10s.​

Answer 1

Answer:

A ) a = v - u / t

= 5 m/s - 0 / 10s

= 0.5 m/s^2

B ) av. speed = u + v / t

= 0 + 5 m/s^2 / 10s

= 0.5 m/s

C) s = ut + 1/2 at^2

= 0 * 10s + 1/2 * 0.5m/s^2 * ( 10s )^2

= 1/2 * 0.5m/s^2 * 100s

= 1/2 * 5/10m/s^2 * 100s

= 1/2 * 50 m

= 25 m

Explanation:

Answer 2

Answers:

• Acceleration of the girl is 0.5 m/s²
• Average speed of the girl is 0.5 m/s
• Distance traveled by the girl in 10 s is 25 m.

Given:

• Initial velocity of the girl = 0 m/s
• Final velocity of the girl = 5 m/s
• Time taken = 10 s

To find:

• Acceleration
• Average speed during 10 s
• Distance she traveled in 10 s

Solution:

We know that

$\boxed {\sf {\red {a= \dfrac{v-u}{t}}}}$

Where,

v denotes Final velocity

u denotes Initial velocity

a denotes Acceleration

t denotes Time

Substitute the values,

$\sf{a = \dfrac{5 - 0}{10} }$

$\sf{a = \dfrac{5}{10} }$

$\boxed {\mathfrak {\green {Acceleration = 0.5\ m/s^{2}}}}$

_______________________________________

We know that,

$\boxed {\sf {\red {Average\ speed = \dfrac{u+v}{t}}}}$

Where,

v denotes Final velocity

u denotes Initial velocity

t denotes Time

Substitute the values,

$\sf {Average\ speed = \dfrac {0+5}{10}}$

$\sf {Average\ speed = \dfrac {5}{10}}$

$\boxed {\mathfrak {\pink {Average\ speed = 0.5\ m/s}}}$

_______________________________________

Using the second equation of motion,

$\boxed {\sf {\red {s=ut+ \dfrac{1}{2}at^{2}}}}$

Where,

v denotes Final velocity

u denotes Initial velocity

a denotes Acceleration

t denotes Time

s denotes distance

Substitute the values,

$\sf {s=0 \times 10+ \dfrac{1}{2} \times 0.5 \times 10^{2}}$

$\sf {s=\dfrac{1}{2} \times 0.5 \times 100}$

$\sf {s=0.5 \times 50}$

$\boxed {\mathfrak {\blue {Distance = 25\ m}}}$