Question

14) A car starting from rest and moves with a velocity 5m/s after 10 seconds. calculate the acceleration and distance travelled by the car during this time​

Answer 1

Answer:

• Acceleration is 0.5 m/s².
• Distance travelled is 25 m.

Given:

• Final velocity (v) = 5 m/s
• Time taken (t) = 10 sec

Explanation:

$\rule{300}{1.5}$

Here, we are given that a car starts from rest [initial velocity (u) = 0 m/s] and moves with the velocity of 5 m/s after 10 sec. We are asked to find the acceleration and the distance travelled by the car during that time.

From first kinematic equation we know,

$\longrightarrow{\sf{v\ =\ u\ +\ at}}$

Substitute the values,

$\longrightarrow{\sf{5\ =\ 0\ +\ a \times 10}} \\ \\ \\ \longrightarrow{\sf{5\ =\ 0\ +\ 10a}} \\ \\ \\ \longrightarrow{\sf{\dfrac{5}{10}\ =\ a}} \\ \\ \\ \longrightarrow{\sf{0.5\ =\ a}} \\ \\ \\ \longrightarrow{\sf{a\ =\ 0.5\ m/s^2}}$

∴ The acceleration of the car is 0.5 m/s².

$\rule{300}{1.5}$

Now, let us find out the distance travelled by the car.

From third kinematic equation we know,

$\longrightarrow{\sf{v^2\ =\ u^2\ +\ 2as}}$

Substitute the values,

$\longrightarrow{\sf{(5)^2\ =\ (0)^2\ +\ 2 \times 0.5 \times s}} \\ \\ \\ \longrightarrow{\sf{25\ =\ 0\ +\ 1s}} \\ \\ \\ \longrightarrow{\sf{\dfrac{25}{1}\ =\ s}} \\ \\ \\ \longrightarrow{\sf{25\ =\ s}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 25\ m}}$

∴ The distance travelled by the car is 25 m.

$\rule{300}{1.5}$