Question

3. A car starting from rest attains a velocity of 60 km/h in 10 minutes. Calculate the
acceleration and the distance travelled by the car, assuming that the acceleration is
uniform.
​

Answer 1

Acceleration $a = 360 \frac{Km}{hr^{2}}$  

Distance travelling by the car $S = \textrm{5 Km}$

Explanation:

Given that,

Initial velocity $u = 0$

Final velocity $v = 60 \frac{Km}{hr}$

Time $t = \textrm{10 min.} = \frac{1}{6} hr$

As we know that, $v =u + at$  

$60 = 0 + a \times \frac{1}{6}$  

Acceleration $a = 60 \times 6 = 360 \frac{Km}{hr^{2}}$  

Assuming that, acceleration is uniform then  

Formula $v^{2} = u^{2} + 2aS$

$60^{2} = 0 + 2\times 360 \times S$

$S = \frac{3600}{ 2\times 360}$

Distance travelling by the car $S = \textrm{5 Km}$.

Answer 2

Acceleration, $a=0.027\ m/s^2$

Distance travelled by the car, d = 5146.09 meters

Explanation:

It is given that,

Initial speed of the car, u = 0

Final speed of the car, v = 60 km/h = 16.67 m/s

Time, t = 10 min = 600 s

Let a is the acceleration of the car. It can be calculated using first equation of motion as :

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

$a=\dfrac{16.67\ m/s}{600\ s}$

$a=0.027\ m/s^2$

So, the acceleration of the car is $0.027\ m/s^2$. Let d is the distance covered by the car. It can be calculated using third equation of motion as :

$v^2-u^2=2ad$

$v^2=2ad$

$d=\dfrac{v^2}{2a}$

$d=\dfrac{(16.67)^2}{2\times 0.027}$

d = 5146.09 meters

Hence, this is the required solution.

Learn more,

Equation of kinematics

https://brainly.in/question/12134828