Question

This is the Question​

Answer 1

Given :

Inital velocity,u 18km/hr = 5m/s

Final velocity ,v = 36km/hr = 10m/s

time ,t = 5sec

To Find :

1) acceleration

2) Distance covered by the car in that time

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

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

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

${\underline{\sf{Answer}}}$

1) We have to Find the acceleration of the car

By equation of motion

$\sf\:v=u+at$

Put the given values

$\sf10=5+5a$

$\sf\:5=5a$

$\sf{a=1ms{}^-2}$

Thus, The acceleration of the car is 1 m/s²

2)We have to find the distance covered by the car in that time .

From the equation of motion.

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

Now put the given values

$\sf\:s=5\times\:5+\frac{1}{2}\times1\times\:{5}^{2}$

$\sf\:s=25+\dfrac{25}{2}$

$\sf\:s=\dfrac{75}{2}$

$\sf\:s=37.5m$

Thus ,the distances covered by the car in that time is 37.5 m.

Answer 2

Given:-

• Initial velocity (u) = 18 km/h
• Final velocity (v) = 36 km/h
• Time taken to accelerate (t) = 5 s

To Find:-

1. Acceleration (a) &
2. Distance travelled while accelerating (s).

Procedure:-

1. (a) :-

We know,

a = (v - u)/t

∴ a = (36 km/h - 18 km/h)/5 s

→ a = (18 km/h)/5 s

→ a = (5 m/s)/5 s

→ a = 1 m/s² ...(Acceleration)

2. (s) :-

We know,

s = ut + ½at²

∴ s = (18 km/h)(5 s) + ½(1 m/s²)(5 s)²

→ s = (5 m/s)(5 s) + (½ m/s²)(25 s²)

→ s = 25 m + (½ m)(25)

→ s = 25 m + 12.5 m

→ s = 37.5 m ...(Distance)