Question

A car accelerates from 6 metre per second to 16 metre per second in 10 second. calculate the acceleration and the distance covered by the car in that time​

Answer 1

Answer:

The acceleration of the car is 1 m/s².

The distance covered by the car is 110 m.

Explanation:

$\bf Given \begin{cases}• & \text{u = 6 m/s.} \\ • & \text{v = 10 m/s.} \\ • & \text{t = 10 sec.}\end{cases}$

To find acceleration of the car = ?

$\qquad {\boxed{\sf V = u + at}}$

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

$\implies \sf a = \dfrac{16-6}{10}$

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

$\implies \sf a = 1 \ m/s^2$

$\therefore$ Acceleration of the car is 1 m/s².

To find distance of the car = ?

$\qquad {\boxed{\sf S = ut + \dfrac{1}{2} at^2}}$

$\implies \sf S = 6 \times 10 + \dfrac{1}{2} \times 1 \times (10)^2$

$\implies \sf S = 60 + \dfrac{1}{2} \times 10 \times 10$

$\implies \sf S = 60 + 50$

$\implies \sf S = 110 \ meter$

$\therefore$ The distance covered by the car is 110 m.

Answer 2

Answer:

The acceleration of the car is 1 m/s².

The distance covered by the car is 110 m.

Explanation:

$\bf Given \begin{cases}• & \text{u = 6 m/s.} \\ • & \text{v = 10 m/s.} \\ • & \text{t = 10 sec.}\end{cases}$

To find acceleration of the car = ?

$\qquad {\boxed{\sf V = u + at}}$

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

$\implies \sf a = \dfrac{16-6}{10}$

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

$\implies \sf a = 1 \ m/s^2$

$\therefore$ Acceleration of the car is 1 m/s².

To find distance of the car = ?

$\qquad {\boxed{\sf S = ut + \dfrac{1}{2} at^2}}$

$\implies \sf S = 6 \times 10 + \dfrac{1}{2} \times 1 \times (10)^2$

$\implies \sf S = 60 + \dfrac{1}{2} \times 10 \times 10$

$\implies \sf S = 60 + 50$

$\implies \sf S = 110 \ meter$

$\therefore$ The distance covered by the car is 110 m.