Question

A car accelerates from 6ms_1to 16ms_1 in 10 sec. Calcuate acceleration anddistance covered by car in that time

Answer 1

ANSWER:

➡acceleration of the car = 1 m/s²

➡distance covered by car in that time = 110 m

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Explanation:

given that,

A car accelerates from 6ms/s to 16m/s in 10 sec,

here,

initial velocity of the car = 6 m/s

and  Similarly ,

final velocity = 10 m/s

given time for acceleration = 10 seconds

now,

1.) Calculate acceleration of the car.

we have,

initial velocity (u) = 6 m/s

final velocity (v) = 16 m/s

time (t) = 10 seconds

by the motion formulae,

v = u + at

where ,

a = acceleration of the car

putting the values,

16 = 6 + a(10)

10a + 6 = 16

10a = 16 - 6

10a = 10

a = 10/10

a = 1 m/s²

so,

acceleration of the car = 1 m/s²

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2.) Calculate distance covered by the car in that time .

now,

we have,

initial velocity (u) = 6 m/s

final velocity (v) = 16 m/s

acceleration (a) = 1 m/s

now,

by the motion formulae

v² = u² + 2as

where,

s = distance coveted in that time,

putting the values,

(16)² = (6)² + 2(1)s

256 = 36 + 2s

2s + 36 = 259

2s = 259 - 39

2s = 220

s = 220/2

s = 110 m

so,

distance covered by car in that time = 110 m

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EXTRA INFORMATION

• When a brake applies in a moving car the acceleration will negative . Negative acceleration is also called Retardation or negative acceleration.

Answer 2

Question:-

A car accelerates from 6m/s to 16m/s in 10 sec. Calcuate acceleration and distance covered by car in that time.

Explanation:

Given, that the car starts from 6 m/s to 16 m/s in 10 seconds.

So,

initial velocity(u)= 6 m/s

final velocity(v)= 16 m/s

time(t)= 10 sec.

♨ Acceleration of the car

✍ Ist. Way

$\rm Now,\ we\ know\ that\ acceleration\ is\ the\ change\ in\ velocity\ in\ a\ specific\ time.$

$\rm So,\ formula\ for\ acceleration\ is\ \frac{v-u}{t}\\ \\ \rm Putting\ values\ in\ the\ equation\\ \\ \implies \frac{16-6}{10}\\ \\ \implies \frac{10}{10}\\ \\ \implies Acceleration= 1 \frac{m}{s^2}$

✍ 2nd. Way

We know the 1st motion of equation, which is v=u+at

We, know the values of v, u and t and need to find a.

So, putting values in the formula

$16=6+a\times (10)\\ \implies 16=6+10a\\ \implies 16-6=10a\\ \implies 10=10a\\ \therefore a=1\frac{m}{s^2}$

♨ Distance covered by the car

Now, we know the initial velocity, final velocity, time and acceleration and we need to find the distance covered by the car.

✍ Way 1

We know the second formula for motion which is $s=ut+\frac{1}{2}at^2\\ \\ \rm Putting\ values\ in\ the\ equation\\ s=6\times 10+\frac{1}{2}\times 1\times 10^2\\ \\ \implies 60+\frac{100}{2}\\ \\ \implies 60+50=110\\ \therefore \rm the\ distance\ covered\ by\ the\ car\ is\ 110\ mts.$

✍ Way 2

We know the 3rd. formula for motion which is $v^2=u^2+2as$

Putting values in the equation,

$(16)^2=(6)^2+2\times 1\times s\\ \implies 256=36+2s\\ \implies 256-36=2s\\ \implies 220=2s\\ \therefore s=110 mts.\\ \rm The\ distance\ covered\ by\ the\ car\ is\ 110\ mts.$

Know what?

Acceleration can also be negative, if initial velocity is greater than the final velocity. And distance can be 0 or positive.