Question

A car accelerates uniformly from 18 km h-¹ to 36 I'm h-¹ in 5s. Calculate:- (I) the Acceleration (II) Distance covered by the car in that time. PLEASE ANSWER, IT IS VERY URGENT

Answer 1

Given,

$\tt u = 18 \: km \ {h}^{ - 1 } = 5 \: m \: {s}^{ - 1}$

$\tt v = 36 \: km \ {h}^{ - 1 } = 10 \: m \: {s}^{ - 1}$

$\tt t = 5 \: s$

(I) Calculating Acceleration

As, we know

$\huge \boxed { \boxed{\tt { a = \frac{v - u}{t}} }}$

$\tt \implies \dfrac{10 \: m \: {s}^{ - 1 } \: - \: 5 \: m \: {s}^{ - 1} }{5 \: s}$

$\tt \implies \dfrac{5 \: m \: {s}^{ - 1} }{5 \: s}$

$\tt \implies 1 \: m \: {s}^{ - 1}$

Thus, acceleration is $\rm 1 \: m \: {s}^{ - 1}$

(II) Calculating Distance Travelled

As, we know

$\huge \boxed { \boxed{\tt { s =ut + \frac{1}{2} \: at ^{2} }}}$

$\tt \implies 5 \: m \: {s}^{ - 1} \times 5 \: s \: + \: \dfrac{1}{2} \times 1 \: m \: {s}^{ - 2} \times (5 \: s) ^{2}$

$\tt \implies 25 \: m \: + \: \dfrac{25}{2} \: m$

$\tt \implies 25 \: m \: + \: 12.5 \: m$

$\tt \implies 37.5 \: m$

Thus, distance travelled is $\rm 37.5 \: m$

Additional Information

• Rate of Change of Distance is known as Speed.

• The speed of body which is applied in beginning is known as Initial Speed and is denoted by u.

• The speed of body which is acquired after the body starts Moving is know as Final Speed and is represented by v.

• Rate of change of Velocity is known as Acceleration.

• Negative Acceleration is known as Retardation.

Important Formulas

$\bf Speed = \dfrac{Distance}{Time}$

$\bf Average \: Speed = \dfrac{Total \: distance}{Total \: time}$

$\bf v = u + at$

$\bf v^{2} - u^{2} = 2as$

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

Answer 2

Explanation:

Acceleration=change in velocity/ time

Change in velocity=36-18=18km/h=5m/s

time=5s

acceleration=5/5=1m/s2

equation of motion,

s=ut+(1/2)at2

u=18 km/h=5 m/s

t=5s

a=1m/s2

s=(5*5)+(1/2*1*5*5)

s=25+12.5 i.e.,=37.5m

formula used:

change in velocity\time=acceleration

s=ut+(1/2) at2