Question

The velocity of a car increases from 30 km per hr to 45 km per hr in 5min .Calculate acceleration and distance travelled​

Answer 1

here is your answer... just using the equations of motion.

Answer 2

GIVEN :-

Initial Velocity of the Car (u) = 30 km/h => 8.33 m/s

Final Velocity of the Car (v) = 45 km/h => 12.50 m/s

Time Taken (t) = 5 Minutes => 300 Seconds

TO FIND :-

The Acceleration (a) of the Car

The Distance Travelled (s) by the Car

SOLUTION :-

(i) Acceleration of the Car -

We can use the Formula, a = $\frac{Final\:Velocity - Initial\:Velocity}{Time\:Taken}$

Substituting the Given Values, we get -

$a = \frac{12.50 - 8.33\:m/s}{300\:Seconds} =>\frac{4.17\:m/s}{300\:seconds} \\\\a = 0.0139\:m/s^2$

∴ The Acceleration (a) of the Car is 0.0139 m/s^2, when Rounded-Off gives 0.014 m/s^2

(ii) The Distance Travelled by the Car -

Using the Third Equation of Motion, (i.e.), $v^2 - u^2 = 2as$, we get -

$\Rightarrow (12.50)^2 - (8.33)^2 = 2 (0.0139)s\\\\\Rightarrow 156.25 - 69.3889 = 0.0278s\\\\\Rightarrow 86.8611 = 0.0278s\\\\Therefore,\:s = \frac{86.8611}{0.0278} = 3124.5m$

∴ The Distance Travelled (s) by the Car is 3124.5 m or 3.1245 km