Question

From rest position, a car moving at uniform acceleration travels a distance of 180m at 12s, what distance will the car cover in 20 seconds?
(I hope my question is clear, My language isn't English so the question can probably be wrong or anything..)

Answer 1

☆Concept:

》car is travelling in uniform acceleration, we can apply the equations of motion.

☆given:

•initial velocity(u)=0 (at rest initially)

☆Formula:

$\boxed{\bf{s=ut+ \dfrac{1}{2}at²}}$

where 's' is the distance covered,'t' is the tame taken.,a is the acceleration.

☆solution:

$\ s=ut+ \dfrac{1}{2}at²$

• $\ 180= 0+ \dfrac{1}{2}a(12)²$

• $\ 180= \dfrac{1}{2}a(12)²$--------(1)

▪︎$\ s' = 0+ \dfrac{1}{2}a(20)²$

▪︎$\ s' = \dfrac{1}{2}a(20)²$--------(2)

Dividing both equations:

》$\dfrac{180}{s'} = \dfrac{ \dfrac{1}{2}a(12)²}{ \dfrac{1}{2}a(20)²}$

》$\dfrac{180}{s'} = \dfrac{ (12)²}{ (20)²}$

》$\dfrac{180}{s'} = \dfrac{144}{400}$

》$\ s' = \dfrac{180×400}{144}$

》$\bf{s' = 500m}$

so the distance travelled in 20s is 500m.

Answer 2

Initial velocity of the car, $\sf u = 0 \: ms^{-1}$

Distance travelled by the car, s = 180 m

Time taken to cover the distance, t = 12 s

By using formula, $\sf s = ut + \dfrac{1}{2} a{t^2}$

$\sf 180 = \dfrac{1}{2} a{(12)^2}$

$\sf 180 = 72a$

$\sf a = \dfrac{180}{72}$

$\sf a = 2.5 \: ms^{-2}$

Now, $\sf s' = \dfrac{1}{2} \: 2.5{(20)^2}$

$\sf s' = 500 \: m$

Therefore, the distance covered by the car in 20 seconds is 500 m.