Question

a race car accelerate on a straight road from rest of 10km/h in 25sec . assume uniform acceleration of the car throughout . find the distance covered in this time?​

Answer 1

Answer:

34.87 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 0 [ As it starts from rest]
• Final velocity (v) = 10 km/h
• Time taken (t) = 25 seconds

We are asked to calculate the distance covered in this time.

Firstly, converting final velocity in its SI unit.

$\longmapsto \rm { v = 10 \; kmh^{-1} }$

• 1 km/ h = 5/18 m/s

$\longmapsto \rm { v =\Bigg ( 10 \times \dfrac{5}{18} \Bigg )\; kmh^{-1} }$

$\longmapsto \rm { v = \dfrac{50}{18} \; kmh^{-1} }$

$\longmapsto \rm { v =2.77 \; kmh^{-1} }$

In order to calculate the distance covered in this time, we need to calculate the acceleration first. By using the first equation of motion :

$\longmapsto \bf { v = u +at}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\longmapsto \rm {2.77 = 0 +25a}$

$\longmapsto \rm {2.77 = 25a}$

$\longmapsto \rm {\dfrac{2.77 }{25}= a}$

$\longmapsto \rm {0.11= a}$

Now, by using the 3rd equation of motion.

$\longmapsto \bf { v^2 -u^2 = 2as}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\longmapsto \rm { (2.77)^2 -(0)^2 = 2(0.11)s}$

$\longmapsto \rm { 7.67 -0= 0.22s}$

$\longmapsto \rm { 7.67 = 0.22s}$

$\longmapsto \rm { \dfrac{7.67}{0.22}=s}$

$\longmapsto \bf { 34.87 \; m \approx s}$

∴ The distance covered in this time is 34.87 metres (approximately).