Question

A racing car has a uniform acceleration of 2m s^2 . what distance will it cover in 10 s after its start?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Distance\:travel=100\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Acceleration(a) = 2 \: m/{s}^{2} \\ \\ \tt: \implies Time(t) = 10 \: s \\ \\ \red{\underline \bold{To \: Find :}}\\ \tt: \implies Distance \:travel(s) = ?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \bold{As \: we \: know \:that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies s =0 \times 10 + \frac{1}{2} \times 2 \times {10}^{2} \\ \\ \tt: \implies s =0 + {10}^{2} \\ \\ \green{\tt: \implies s = 100 \: m} \\ \\ \bold{Alternate \: method : } \\ \tt : \implies v = u + at \\ \\ \tt : \implies v = 0 + 2 \times 10 \\ \\ \green{\tt : \implies v = 20 \: m/s }\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {20}^{2} = {0}^{2} + 2 \times 2 \times s \\ \\ \tt: \implies 400 = 4 \times s \\ \\ \tt: \implies s = \frac{400}{4} \\ \\ \green{\tt: \implies s = 100m}\\\\ \green{\tt \therefore Distance \: travelled \: by \: racing \: car \: is \: 100 \: m} \\ \\ \blue{ \bold{Some \: related \: formula : }} \\ \orange{\tt \circ \: v = u + at} \\ \\ \orange{\tt \circ \: {v}^{2} = {u}^{2} + 2as}$

Answer 2

Answer:-

Given:

acceleration (a) = 2 m/s²

Time (t) = 10 s

As it starts from rest , Initial velocity (u) = 0 m/s

We know that,

S (Distance) = u*t + 1/2 * a * t²

S = (0)(10) + 1/2 * (2) * (10)²

S = 0 + 100

S = 100 m.

(or)

We know that,

v (Final Velocity) = u + a*t

v = 0 + (2)(10)

v = 20 m/s

Again,

We know that,

v² - u² = 2*a*S

(20)² - (0)² = (2)(2)(S)

400 - 0 = 4S

4S = 400

S = 400/4

S = 100 m

Hence,it covers 100 m in 10 s after its start.