Question

A car is traveling along a straight road with a velocity of 10 m/s. It begins to accelerate uniformly at time t = 0 and covers a distance of 300 m in 5 s. What is the magnitude of the acceleration?

Answer 1

The position is given by:

$X_{f}=X_{o}+V_{ox}t+\frac{1}{2}a_{x}t^{2}$

Where

• The initial position is

$X_{i}=0m$

• The final position is

$X_{f}=300m$

• The initial speed is

$V_{ox}=10\frac{m}{s}$

• Acceleration

$a_{x}$

• Time is

$t=5.00s$

Expression for acceleration

$X_{f}=X_{o}+V_{ox}t+\frac{1}{2}a_{x}t^{2}\\ X_{f}-X_{o}-V_{ox}t=\frac{1}{2}a_{x}t^{2}\\ \frac{1}{2}a_{x}t^{2}=X_{f}-X_{o}-V_{ox}t\\ a_{x}=\frac{2(X_{f}-X_{o}-V_{ox)t}}{t^{2}}$

Numerically evaluating

$a_{x}=\frac{2(300m-0m-10\frac{m}{s}*5.00s}{(5.0s)^{2}}$

FInally

$a_{x}=20\frac{m}{s^{2}}$

The acceleration of the car is

$\boxed{a_{x}=20\frac{m}{s^{2}}}$