Question

If p(y)=y^3-5y^2+5y+10 then p(2) + p(-2)=?​

Answer 1

Answer:

A car accelerates uniformly from 30 km/h to 60 km/h in 5 seconds Calculate ::

(i) Acceleration in m/s²?

(ii) Distance covered by the car in metres during this interval?

Answer:

(i) Acceleration of car in m/s² is 1.66 m/s².

(ii) Distance covered by the car in metres during this interval is 62.4 meters.

Explanation:

Given that:

Initial velocity (u) = 30 km/h = (30 × 5/18) m/s = 8.33 m/s

Final velocity (v) = 60 km/h = (60 × 5/18) m/s = 16.66 m/s

Tine taken (t) = 5 seconds

To Find:

Acceleration (a)?

Distance covered (s)?

Solution:

Using first equation of motion ::

We know that,

⋆ \boxed{\bf{v = u + at}}

v=u+at

According to the question by using the formula we get,

➻ \sf 16.66 = 8.33 + (a\:\times\:5)16.66=8.33+(a×5)

➻ \sf 16.66 - 8.33 = 5a16.66−8.33=5a

➻ \sf 5a = 8.335a=8.33

➻ \sf a = {\cancel{\dfrac{8.33}{5}}}a=

5

8.33

➻ \bf\red{a = 1.66\:m/s^2}a=1.66m/s

2

∴ Hence, acceleration of car in m/s² is 1.66 m/s².

Now, let's find distance covered by the car. Using second equation of motion ::

We know that,

⋆ \boxed{\bf{s = ut + \dfrac{1}{2}at^2}}

s=ut+

2

1

at

2

According to the question by using the formula we get,

➻ \sf s = (8.33\:\times\:5) + \dfrac{1}{\cancel{2}}\:\times\:\cancel{1.66}\:\times\:(5)^2s=(8.33×5)+

2

1

×

1.66

×(5)

2

➻ \sf s = 41.65 + (0.83\:\times\:35)s=41.65+(0.83×35)

➻ \sf s = 41.65 + 20.75s=41.65+20.75

➻ \bf\purple{s = 62.4\:m}s=62.4m

∴ Distance covered by the car in metres during this interval is 62.4 meters.

Know More:

\clubsuit♣ Three equations of motion ::

v = u + atㅤㅤ ㅤ ㅤ [Used above]

s = ut + ½ at²ㅤㅤㅤ[Used above]

v² = u² + 2as

\clubsuit♣ Some important definitions ::

Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.