Question

8.
A car, initially at rest, picks up a velocity of 72 kmh-l in 1/4 minute. What is the distance
travelled by the car?
A) 180 m
B) 160 m
C) 120 m
D) 150 m
If you answer, I will mark you as the brainliest ​

Answer 1

${\huge{\boxed{\mathcal{\red{Answer}}}}}$

Option D) 150m

${\huge{\boxed{\mathcal{\red{Explanation}}}}}$

It is given that :-

Initial velocity of car = 0m/s (since it was at rest)

Final velocity = 72km/h

$= 72 \times \frac{5}{18} m/s$

= 20m/s.

Time taken to reach final velocity = 1/4 min.

$\frac{1}{4} \times 60$

$= 15sec$

Now if we talk about the acceleration, then

acceleration is the rate of change velocity, thus

$a = \frac{v - u}{t}$

After inputting the values in this equation we get :

$a = \frac{20 - 0}{15}$

$a = 1.33m/ {s}^{2}$

Now according to second equation of motion the distance traveled by car is :

$s = ut + \frac{1}{2} \times a {t}^{2}$

$s = 0 + \frac{1}{2} \times 1.33 \times {15}^{2}$

$s = \frac{1}{2} \times 1.33 \times 225$

$s = 1.33 \times 112.5$

$s = 150m$

Therefore the distance travelled by car is 150m.

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BY SCIVIBHANSHU

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