Question

20) Deepa accelerates her scooty uniformly from 18km/h to 45km/h in 5 second.

Calculate:

a) the acceleration

b) the distance covered by her in that time​

Answer 1

Question :

Deepa accelerates her scooty uniformly from 18km/h to 45km/h in 5 second.

Calculate:

a) the acceleration

b) the distance covered by her in that time

To find :

a) the acceleration

b) the distance covered by her in that time

Given Data :

$\sf \small \: Initial \: Velocity (u) = 18 \: km/h = 5m/sec$

$\sf \small \: Final \: Velocity (v) = 45 km/h = \frac{25}{2} m/sec$

$\sf \: Time = 5 \: seconds$

Formula Used :

For acceleration we will use Newton's 1st law of motion which states :

$\sf \: a= \frac{v - u}{t}$

For getting distance we need to apply Newton's 2nd law of motion which states :

$\sf \: S = ut + \frac{1}{2} a {t}^{2}$

Solution :

$\sf \small \: a) \: Acceleration$

$\sf \small \: Applying \: a= \frac{v - u}{t} \: \: we \: get$

$\sf \small \rightarrow \: a = \frac{ \frac{25}{2} - 5 }{5}$

$\sf \small \rightarrow a = \frac{ \frac{25 - 10}{2} }{5}$

$\sf \small \rightarrow \: a = \frac{ \frac{15}{2} }{5}$

$\boxed{\sf \small \: a = \frac{3}{2} m \: {sec}^{ - 2} \: or \: 1.5m \: {sec}^{ - 2}}$

Now the second part we need to get the distance using the second law of motion :

$\sf \small \: b) \: distance$

$\sf \small \rightarrow \:S = ut + \frac{1}{2} a {t}^{2}$

$\sf \small \rightarrow \: S = 5 \times 5+ \frac{1}{2} \times \frac{3}{2} \times {(5)}^{2}$

$\sf \small \rightarrow \: S = 25 + \frac{1}{2} \times \frac{3}{2} \times 25$

$\sf \small \rightarrow \: S = 25 + \frac{3}{4} \times 25$

$\sf \small \rightarrow \: S = 25 + \frac{75}{4}$

$\sf \small \rightarrow \: S = \frac{100 + 75}{4}$

$\sf \small \rightarrow \: S = \frac{175}{4} m$

$\boxed{\sf \small \: S = 43.75m}$

Final Answer :

• Acceleration is 1.5 m/sec²

• Distance is 43.75 m

Answer 2

Step-by-step explanation:

Formula Used :

For acceleration we will use Newton's 1st law of motion which states :

\sf \: a= \frac{v - u}{t}