Question

A car starts from rest and accelerates uniformly over a time of 6.45 seconds for a distance of 300 m. Determine the acceleration of the car.

Answer 1

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

$\green{\tt{\therefore{Acceleration=14.4\:m/s^{2}}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Distance(s) = 300\:m \\ \\ \tt: \implies Time(t) = 6.45\: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Acceleration(a) = ?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \tt \circ \: Time = 6.45 \: sec \\ \\ \tt \circ \: Distance = 300 \: m \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \tt: \implies 300 = 0 \times 6.45 + \frac{1}{2} \times a \times {6.45}^{2} \\ \\ \tt: \implies 300 = 0 + \frac{1}{2} \times a \times 41.6025 \\ \\ \tt: \implies 300 \times 2 = 41.6025 \times a \\ \\ \tt: \implies a = \frac{600}{41.6025} \\ \\ \green{\tt: \implies a = 14.4 \: {m/s}^{2} }$

Answer 2

Given:-

A car starts from rest and accelerates uniformly over a time of 6.45 seconds for a distance of 300 m.

To find:-

Acceleration of the car.

Solution:-

Initial velocity(u) = 0 m/s

(As the car was in rest initially)

Time(t) = 6.45 s

Distance(s) = 300 m

Using second equation of motion,

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

$\bold{300=0*6.45+\frac{1}{2}a*(6.45)^2}$

$\bold{300=0+\frac{1}{2}*a*41.6025}$

$\bold{300*2=41.6025*a}$

$\bold{600/41.6025=a}$

$\boxed{\bold{a=14.42\ m/s^2}}$

∴So, the acceleration of the car is 14.42 m/s².