Question

16. A driver decreases the speed of a car from 30 m/s to 15 m/s in 3 seconds.
Find the acceleration of the car.​

Answer 1

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First of all we should note that in this problem the term 'speed' is being used in the same sense as 'velocity'. <br> Here,

Initial velocity of car, u = 25 m/s <br>

Final velocity of car, v = 10 m/s <br> And,

Time taken, t = 5s <br> Now, putting these values in the formula for acceleration : <br>

<br> We get:

<br>

<br>

<br> Thus, the acceleration of car is,

(minus 3

). The negative sign of acceleration means that it is retardation. So, we can also say that the car has a retardation of

.

Answer 2

Given:

• Initial velocity of car (u) = 30m/s
• Final velocity of car (v) = 15m/s
• Time taken (t) = 3 second

To Find:

• Acceleration of car ?

Solution:

We have given that initial velocity and final velocity of car is 30m/s and 15m/s respectively and time taken to decrease velocity is 3 second.

We know that :

• Acceleration = $\large{\sf{\dfrac{v-u}{t}}}$

Where,

• v = final velocity
• u = initial velocity
• t = time taken

Put values in formula:

$\large{\sf{Acceleration~=~\dfrac{v-u}{t}}}$

$\implies\large{\sf{Acceleration~=~\dfrac{15-30}{3}}}$

$\implies\large{\sf{Acceleration~=~\dfrac{-15}{3}}}$

$\implies\large{\sf{Acceleration~=~\cancel\dfrac{-15}{3}}}$

$\implies\large{\sf{Acceleration~=~ -5}}$

Therefore,

• $\huge{\boxed{\large{\bf{\pink{Acceleration~of~car~is~{\underline{-5m/s²}}}}}}}$