To find the acceleration of the car while stopping, we can use the equation of motion:
v=u+atv = u + atv=u+at
where:
- vvv is the final velocity (which is 0 m/s since the car comes to a stop).
- uuu is the initial velocity (32 m/s to the left).
- aaa is the acceleration.
- ttt is the time taken to come to a stop (5 seconds).
Plugging in the given values:
0 m/s=32 m/s+a⋅5 s0 , ext{m/s} = 32 , ext{m/s} + a cdot 5 , ext{s}0m/s=32m/s+a⋅5s
Now, let's solve for aaa:
a⋅5 s=−32 m/sa cdot 5 , ext{s} = -32 , ext{m/s}a⋅5s=−32m/s
a=−32 m/s5 sa = frac{ -32 , ext{m/s}}{5 , ext{s}}a=5s−32m/s
a=−6.4 m/s2a = -6.4 , ext{m/s}^2a=−6.4m/s2
So, the acceleration of the car while stopping is −6.4 m/s2-6.4 , ext{m/s}^2−6.4m/s2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which is necessary to bring the car to a stop.