Question

A car of mass 1000 kg develops a force of 500 N over a distance of 49 m. If initially the car was at rest the time for which it accelerates is?

Answer 1

Given :

▪ Mass of car = 1000kg

▪ Force acts on car = 500N

▪ Distance covered = 49m

▪ Initial velocity of car = zero

To Find :

▪ Time of acceleratory period.

Calculation :

⚾ Acceleration of car :

$\implies\bf\:F=ma\\ \\ \implies\sf\:500=1000a\\ \\ \implies\sf\:a=\dfrac{500}{1000}\\ \\ \implies\boxed{\bf{\purple{a=0.5\:ms^{-2}}}}$

⚾ Time interval :

$\mapsto\sf\:s=ut+\dfrac{1}{2}at^2\\ \\ \mapsto\sf\:49=(0\times t)+\dfrac{1}{2}(0.5\times t^2)\\ \\ \mapsto\sf\:49=0.25t^2\\ \\ \mapsto\sf\:t^2=\dfrac{49}{0.25}\\ \\ \mapsto\sf\:t^2=196\\ \\ \mapsto\underline{\boxed{\bf{\orange{t=14\:s}}}}$

Answer 2

Answer:

14 sec

Explanation:

Given :

Mass m = 1000 kg

Force F = 500 N

a = F / m

= > a = 500 / 1000 = 0.5 m / sec²

Initial velocity u = 0 m / sec

Distance s = 49 m :

Using second equation of motion :

s = u t + 1 / 2 a t²

= > 49 = 0 + 1 / 2 × 0.5 × t²

= > t² = 49 / 0.25

= > t = ± 7 / 0.5

Since time can't be negative :

Therefore , required time is ( 7 / 0.5 ) = 14 sec!