Question

A car, initially at rest, picks up a velocity of 72 km h$^{-1}$ in 1/4 minute. Calculate acceleration and distance covered by the car​

Answer 1

Answer

Given :-

• Initial velocity, u = 0 km/hr = 0 m/s
• Final velocity, v = 72 km/hr = 20 m/s
• Time taken, t = 1/4 min = 15 sec

To find :-

• Acceleration - a
• Distance travelled - s

Solution :-

Calculating Acceleration :-

• u = 0 m/s
• v = 20 m/s
• t = 15 sec

Substituting the value in 1st equation of motion :-

$\sf v = u + at$

⟹ $\sf 20 = 0 + 15a$

⟹ $\sf 15a = 20$

⟹ $\sf a = \frac{20}{15}$

⟹ $\sf a = 1.33 m/s^2$

$\boxed{\underline{\bf\red{ Acceleration = 1.33 m/s^2 }}}$

Calculating distance travelled :-

• u = 0 m/s
• a = 1.33 m/s²
• t = 15 sec

Substituting the value in 2nd equation of motion :-

$\sf s = ut + \frac{1}{2} at^2$

⟹ $\sf s = 0 + \frac{1}{2} \times 1.33 \times 15^2$

⟹ $\sf s = \frac{1}{2} \times 300$

⟹ $\sf s = 150 m$

$\boxed{\underline{\bf\red{Distance\: Travelled = 150 m }}}$

Answer 2

Answer:

Given:-

u = 0

v = 72 km h$^{-1}$

t = 1/4 minute = 15 s.

where u means initial velocity

v means final velocity and

t means time.

To find:-

Acceleration and distance covered by the car..

Solution:-

Using first equation of motion

v = u + at

Substitute the known values

$\longrightarrow$ 20 = 0 + a × 15

$\longrightarrow$ a = 20/15 = 4/3

= 1.33 ms$^{–2}$

Thus, acceleration = 1.33 ms$^{–2}$

Now using second equation of motion

S = ut + 1/2at$^{2}$.

$\longrightarrow$ S = 0 × 30 + 1/2 × 4/3 × (15)$^{2}$.

= 150 m

Therefore, acceleration is 1.33 ms$^{–2}$ and distance covered by car is 150 m.

Equations of motion:-

First equation of motion:

• v = u + at

Second equation of motion

• S = ut + 1/2at$^{2}$

Third equation of motion

• v$^{2}$ – u$^{2}$ = 2as