could someone please help with this physics problem 😭

[hard hound]

I feel like this is really simple but I don’t know how to do it..

[dry vigil]

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[mint olive]

Use v = U + At

And s= ut + 1/2 at²

[weak crescent]

divide the total distance into 3 parts:
a = distance covered while the car accelerates
b = distance covered while speed is constant
c = distance covered while car decelerates

maximum speed would be the speed of the car when it's speed was constant, or it's final speed after undergoing acceleration

it accelerated at the rate of 2ms^(-2) for 6s ryt?

so use
v = u + at
you have a=2ms^(-2) , t=6s , u=0ms^(-1) (car starts from rest)

and for the distance a just use s=ut+1/2 at^2
you have u, a and t

distance b is just whatever speed you got in a * 30s

for distance c

you have u as whatever speed you got in a
v as 0
t as 5

and you know $a = \frac{v-u}{t}$

[forest vortex]

Inverse Cupid

[weak crescent]

divide the total distance into 3 parts:

a = distance covered while the car accelerates

b = distance covered while speed is constant

c = distance covered while car decelerates

maximum speed would be the speed of the car
when it's speed was constant, or it's final speed after undergoing acceleration

it accelerated at the rate of $2ms^{-2}$ for 6s

so use

v = u + at

you have $a=2ms^{-2}$ , t=6s , $u=0ms^{-1}$
(car starts from rest)

and for the distance a just use $s=ut+\frac{at^2}{2}$

you have u, a and t

distance b is just whatever speed you got in a * 30s

for distance c

you have u as whatever speed you got in a

v as 0

t as 5

and you know $a = \frac{v-u}{t}$

then $s=ut+\frac{at^2}{2}$ again to find distance c

[forest vortex]

Inverse Cupid

[hard hound]

thanks so much!!