Question

Find the acceleration of a car and its displacement from the following observations
Time speed (in km/h)
3:15pm 26
3:45pm 60

Answer 1

Given that,

Initial velocity =26 km/hr = 7.2 m/s

Final velocity =60 km/hr = 16.6 m/s

Time 3:15 pm to 3:45 pm.

We need to calculate the acceleration of the car

Using formula of acceleration

$a=\dfrac{v_{2}-v_{1}}{t_{2}-t_{1}}$

Where, v₂₁ = initial velocity

v₂ = final velocity

t = time

Put the value into the formula

$a=\dfrac{16.6-7.2}{1800}$

$a=0.0052\ m/s^2$

We need to calculate the displacement

Using equation of motion

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

Put the value into the formula

$s=\dfrac{16.6^2-7.2^2}{2\times0.0052}$

$s=21511.5\ m$

$s=21.5\ km$

The displacement is 21.5 km.

Hence, The acceleration is 0.0052 m/s².

The displacement is 21.5 km.

Answer 2

Given that,

Initial velocity =26 km/hr = 7.2 m/s

Final velocity =60 km/hr = 16.6 m/s

Time 3:15 pm to 3:45 pm.

We need to calculate the acceleration of the car

Using formula of acceleration

$a=\dfrac{v_{2}-v_{1}}{t_{2}-t_{1}}$

Where, v₂₁ = initial velocity

v₂ = final velocity

t = time

Put the value into the formula

$a=\dfrac{16.6-7.2}{1800}$

$a=0.0052\ m/s^2$

We need to calculate the displacement

Using equation of motion

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

Put the value into the formula

$s=\dfrac{16.6^2-7.2^2}{2\times0.0052}$

$s=21511.5\ m$

$s=21.5\ km$

The displacement is 21.5 km.

Hence, The acceleration is 0.0052 m/s².

The displacement is 21.5 km.