Question

a car acquires a velocity of 72 km per hour in 10 second starting from rest find the acceleration average velocity and distance travelled in this time

Answer 1

hey the answer is,
weight of car =72 km =72×1000m=72000
time=10 sec
average distance=72000×10=720000

Answer 2

Correct Question:

A car acquires a velocity of 72 kilometre per hour in 10 seconds starting from rest. Find the acceleration, average velocity and distance travelled in this time.

Answer:

• The acceleration of the car is 2 m/s².

• The average velocity of the car is 10 m/s.

• The distance travelled in this time by the car is 100m.

Explanation:

Given:

Initial velocity of the car (u) = 0

Final velocity of the car (v) = 72 km/h

Time taken (t) = 10 seconds

Since the time is in seconds we have to convert the distance into metre.

So,

As we know,

1 km = 1000 m and

$\frac{72 \times 1000m}{60 \times 60s} = 20 m/s$

Now,

$\huge{\underline{\pink{\tt{Acceleration \: of \: the\: car:}}}}$

Solution :

Acceleration of the car (a) = $\frac{v-u}{t}$

$= \frac{(20-0) m/s}{10 s} = 2 m/{s}^{2}$

Therefore, the acceleration of the car is 2 m/s².

$\huge{\underline{\blue{\tt{Average \: velocity \: of \: the \: car :}}}}$

Average velocity = $\frac{u+v}{2}$

$= \frac{0+20}{2} m/s = 10 m/s$

Therefore, the average velocity of the car is 10 m/s.

$\huge{\underline{\pink{\tt{Distance \: travelled \: by \: the \: car:}}}}}$

Distance travelled (S) = $ut+ \frac{1}{2} a{t}^{2}$

$0 + \frac{1}{2} \times 2 \times {10}^{2}$

= 100 metres.

Therefore, the distance travelled by the car is 100 m.