Calculating time taken and distance

Question

A car accelerates from $10 \text{ m/s}$ to $ 60 \text{ m/s}$ at an acceleration of $10 \text{ m/s$^2$}$. Calculate the time taken for this acceleration to occur and the distance the car has traveled in time.

I know that the acceleration is $10 \text{ m/s$^2$}$ and the initial velocity is $10 \text{ m/s}$ and the final velocity is $60 \text{ m/s}$.

However, I am unsure whether to use the equation $v = u + at$ or the equation $s = ut + \frac {1}{2}at^2$.

Answer 1

You were on the right track. You know that $u=10$, $v=60$, $a=10$

In the equation v=u+at the only thing you're missing is time so that formula can be used to get the time taken.

The other formula will be useful once you know the time taken.

Answer 2

v = u + at is the equation you will want to use.

Answer 3

1) to calculate time t use first law of motion, $v=u+at$, $$60=10+10t$$ $$t=\frac{50}{10}=5 sec$$ 2) to calculate distance s traveled in time $t=5sec$ use 2nd law of motion, $s=ut+\frac12 at^2$, $$s=10\cdot 5+\frac12 \cdot 10\cdot 5^2=175 m$$